5 Marks Questions - Physics STD 12 Science Questions - Vidyadip

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Question 15 Marks

What is the focal length of a convex lens of focal length 30cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

Answer

Focal length of the convex lens, $f_1 = 30$ cm
Focal length of the concave lens, $f_2 = - 20$ cm
Focal length of the system of lenses = f
The equivalent focal length of a system of two lenses in contact is given as:
$\frac{1}{\text{f}}=\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}$
$\frac{1}{\text{f}}=\frac{1}{30}-\frac{1}{20}=\frac{2-3}{60}=-\frac{1}{60}$
$\therefore \ \text{f}=-60 \ \text{cm}$
Hence, the focal length of the combination of lenses is 60 cm. The negative sign indicates that the system of lenses acts as a diverging lens.

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Question 25 Marks

Determine the ‘effective focal length’ of the combination of the two lenses in if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all.

Answer

Here, Focal length of first lens, $f_1 = 30$ cm
Focal length of second lens, $f_2 = -20$ cm
Distance between the lens, $d = 8.0$ cm
Let a parallel beam be incident on the convex lens first. If second lens were absent, then
$\text{u}_1=\infty \ \text{and} \ \text{f}_1=30 \ \text{cm}$
In the formula, as
$\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}_1}$
$\therefore \ \frac{1}{\text{v}_1}-\frac{1}{\infty}=\frac{1}{30}$
i.e.$, v_1 = 30\ cm$
This image would now act as a virtual object for second lens.
$\therefore \ \text{u}_2=+(30-8)=+22 \ \text{cm}$
$f_2 = -20 cm$
Since,
$\frac{1}{\text{v}_2}=\frac{1}{\text{f}_2}+\frac{1}{\text{u}_2}$
$\frac{1}{\text{v}_2}=\frac{1}{-20}+\frac{1}{22}$
$=\frac{-11+10}{220}=\frac{-1}{220}$
$\therefore$ Parallel incident beam would appear to diverge from a point 220 - 4 = 216 cm from the centre of the two lens system.
Assume that a parallel beam of light from the left is incident first on the concave lens.
$\therefore \ \text{u}_1=-\infty \ \text{and} \ \text{f}_1=-20 \ \text{cm}$
As, $\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}_1}$
$\therefore$ $\frac{1}{\text{v}_1}=\frac{1}{\text{f}_1}+\frac{1}{\text{u}_1}$
$=\frac{1}{-20}+\frac{1}{-\infty}$
$=\frac{1}{-20}$
$v_1 = -20$ cm
This image acts as a real object for the second lens
$u_2= -(20 + 8) = -28$ cm and,
$f_2= 30$ cm
Since, $\frac{1}{\text{v}_2}-\frac{1}{\text{u}_2}=\frac{1}{\text{f}_2}$
$\therefore \ \frac{1}{\text{v}_2}=\frac{1}{\text{f}_2}+\frac{1}{\text{u}_2}$
$=\frac{1}{30}-\frac{1}{28}$
$=\frac{14-15}{420}$
$\text{v}_2=-420 \ \text{cm}$
$\therefore$ The parallel beam appears to diverge from a point 420 - 4 = 416 cm, on the left of the centre of the two lens system.
We finally conclude that the answer depends on the side of the lens system where the parallel beam is incident. Therefore, the notion of effective focal length does not seem to be meaningful here.

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Question 35 Marks

A Cassegrain telescope uses two mirrors. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?

Answer

The following figure shows a Cassegrain telescope consisting of a concave mirror and a convex mirror.
Distance between the objective mirror and the secondary mirror, d = 20 mm

Radius of curvature of the objective mirror, $R_1= 220$ mm
Hence, focal length of the objective mirror, $\text{f}_1=\frac{\text{R}_1}{2}=110\ \text{mm}$
Radius of curvature of the secondary mirror, $R_2 = 140$ mm
Hence, focal length of the secondary mirror, $\text{f}_2=\frac{\text{R}_2}{2}=\frac{140}{2}=70\ \text{mm}$
The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for the secondary mirror.
Hence, the virtual object distance for the secondary mirror, $u = f_1- d$
$= 110 - 20$
$= 90$ mm
Applying the mirror formula for the secondary mirror, we can calculate image distance (v)as:
$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}_2}$
$\frac{1}{\text{v}}=\frac{1}{\text{f}_2}-\frac{1}{\text{u}}$
$=\frac{1}{70}-\frac{1}{90}=\frac{9-7}{630}=\frac{2}{630}$
$\therefore \text{v}=\frac{630}{2}=315 \ \text{mm}$
Hence, the final image will be formed 315 mm away from the secondary mirror.

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Question 45 Marks

An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?

Answer

Focal length of the objective lens, $f_0 = 1.25$ cm
Focal length of the eyepiece, $f_e = 5$ cm
Least distance of distinct vision, $d = 25$ cm
Angular magnification of the compound microscope = 30X
Total magnifying power of the compound microscope, m = 30
The angular magnification of the eyepiece is given by the relation:
$\text{m}_\text{e}=\Big(1+\frac{\text{d}}{\text{f}_\text{e}}\Big)$
$=\Big(1+\frac{25}{5}\Big)=6$
The angular magnification of the objective lens (mo) is related to meas:
$m_e m_e= m$
$\text{m}_0=\frac{\text{m}}{\text{m}_\text{e}}$
$=\frac{30}{6}=5$
We also have the relation:
$\text{m}_0=\frac{\text{image distance for the objective lenc(v}_0)}{\text{image distance for the objective lenc(-u}_0)}$
$5=\frac{\text{v}_0}{\text{-u}_0}$
$\therefore \ \text{v}_0=-5\text{u}_0$
Applying the lens formula for the objective lens.
$\frac{1}{\text{f}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{u}_0}$
$\frac{1}{1.25}=\frac{1}{-5\text{u}_0}-\frac{1}{\text{u}_0}=\frac{-6}{5\text{u}_0}$
$\therefore \ \text{u}_0=\frac{-6}{5}\times1.25=-1.5 \ \text{cm}$
And, $\text{v}_0=-5\text{u}_0$
$= -5 × (-1.5) = 7.5$ cm
The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.
Applying the lens formula for the eyepiece:
$\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{f}_\text{e}}$
Where, $v_e =$ Image distance for the eyepiece $= -d= -25$ cm
$u_e =$ Object distance for the eyepiece
$\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{f}_\text{e}}$
$=-\frac{1}{25}-\frac{1}{5}=\frac{-6}{25}$
$\therefore \ \text{u}_\text{e}=-04.17 \ \text{cm}$
Separation between the objective lens and the eyepiece $=|\text{u}_\text{e}|+|\text{v}_0|$
$= 4.17 + 7.5$
$ = 11.67$ cm
Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.

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Question 55 Marks

Use the mirror equation to deduce that:

an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
a convex mirror always produces a virtual image independent of the location of the object.
the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.

Answer

For a concave mirror, the focal length (f) is negative.

$\therefore \ \text{f}<0$

When the object is placed on the left side of the mirror, the object distance (u)is negative.

$\therefore \ \text{u}<0$

For image distance v, we can write the lens formula as:

$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}\dots(1)$

The object lies between f and 2f.

$\therefore \ 2\text{f}<\text{u}<\text{f}$ $(\therefore \ \text{u and f are negetive})$

$\frac{1}{2\text{f}}>\frac{1}{\text{u}}>\frac{1}{\text{f}}$

$-\frac{1}{2\text{f}}>-\frac{1}{\text{u}}>-\frac{1}{\text{f}}$

$\frac{1}{\text{f}}-\frac{1}{2\text{f}}<\frac{1}{\text{f}}-\frac{1}{\text{u}}<0\dots(2)$

Using equation (1), we get:

$\frac{1}{2\text{f}}<\frac{1}{\text{v}}<0$

$\therefore \ \frac{1}{\text{v}}$ is negative, i.e., v is negative.

$\frac{1}{2\text{f}}<\frac{1}{\text{v}}$

2f > v

-v > -2f

Therefore, the image lies beyond 2f.

For a convex mirror, the focal length (f) is positive.

$\therefore \ \text{f}>0$

When the object is placed on the left side of the mirror, the object distance (u) is negative.

$\therefore \ \text{u}>0$

For image distance v, we have the mirror formula:

$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}$

Using equation (2), we can conclude that:

$\frac{1}{\text{v}}<0$

v > 0

Thus, the image is formed on the back side of the mirror.

Hence, a convex mirror always produces a virtual image, regardless of the object distance.

For a convex mirror, the focal length (f) is positive.

$\therefore \ \text{f}>0$

When the object is placed on the left side of the mirror, the object distance (u) is negative,

$\therefore \ \text{u}>0$

For image distance v, we have the mirror formula:

$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}$

But we have u < 0

$\therefore \ \frac{1}{\text{v}}>\frac{1}{\text{f}}$

v < f

Hence, the image formed is diminished and is located between the focus (f) and the pole.

For a concave mirror, the focal length (f) is negative

$\therefore \ \text{f}>0$

When the object is placed on the left side of the mirror, the object distance (u) is negative.

$\therefore \ \text{u}>0$

It is placed between the focus (f) and the pole.

$\therefore \ \text{f}>\text{u}>0$

$\frac{1}{\text{f}}<\frac{1}{\text{u}}<0$

$\frac{1}{\text{f}}-\frac{1}{\text{u}}<0$

For image distance v, we have the mirror formula:

$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}$

$\therefore \ \frac{1}{\text{v}}<0$

v > 0

The image is formed on the right side of the mirror. Hence, it is a virtual image. For u < 0 and v > 0, we can write:

$\frac{1}{\text{u}}>\frac{1}{\text{v}}$

v > u

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Question 65 Marks

a cross - section of a ‘light pipe’ made of a glass fibre of refractive. The outer covering of the pipe is made of a material of refractive. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.
What is the answer if there is no outer covering of the pipe?

Answer

Refractive index of the glass fibre, $µ_1 = 1.68$

Refractive index of the outer covering of the pipe, $µ_2 = 1.44$
Angle of incidence = i
Angle of refraction = r
Angle of incidence at the interface = i'
The refractive index (µ) of the inner core - outer core interface is given as:
$\mu=\frac{\mu_2}{\mu_1}=\frac{1}{\sin\text{i}'}$
$\sin\text{i}'=\frac{\mu_1}{\mu_2}$
$=\frac{1.44}{1.68}=0.8571$
$\therefore \ \text{i}'=59^\circ$
For the critical angle, total internal reflection (TIR) takes place only when i > i, i. e., i > 59°
Maximum angle of reflection, $r_{max} = 90^\circ - i' = 90^\circ - 59^\circ = 31^\circ$
Let, $i_{max}$ be the maximum angle of incidence.
The refractive index at the air - glass interface, $µ_1 = 1.68$
We have the relation for the maximum angles of incidence and reflection as:
$\mu_1=\frac{\sin\text{i}_\text{max}}{\sin\text{r}_\text{max}}$
$=\sin\text{i}_\text{max}=\mu_1\sin\text{r}_\text{max}$
$= 1.68 sin 31°$
$= 1.68 × 0.5150$
$= 0.8652$
$\text{i}_\text{max}=\sin^{-1}0.8652\approx60^\circ$
Thus, all the rays incident at angles lying in the range 0 < i< 60° will suffer total internal reflection.

If the outer covering of the pipe is not present, then:

Refractive index of the outer pipe, $µ_1 =$ Refractive index of air = 1
For the angle of incidence i = 90°, we can write Snell's law at the air - pipe interface as:
$\frac{\sin\text{i}}{\sin\text{r}}=\mu_2=1.68$
$\sin\text{r}=\frac{\sin90^\circ}{1.68}=\frac{1}{1.68}$
$\text{r}=\sin^{-1}(0.5952)$
= 36.5°
$\therefore \ \text{i}'=90^\circ-36.5^\circ=53.5^\circ$
Since i' > r, All incident rays will suffer total internal reflection.

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Question 75 Marks

A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Answer

Height of the needle, $h_1 = 4.5$ cm
Object distance, $u = – 12$ cm
Focal length of the convex mirror, $f = 15$ cm
Image distance $= v$
The value of v can be obtained using the mirror formula:
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\frac{1}{\text{u}}=\frac{1}{\text{v}}-\frac{1}{\text{f}}$
$=\frac{1}{-25}-\frac{1}{5}=\frac{-5-1}{25}=\frac{-6}{25}$
$\therefore \ \text{u}=-\frac{25}{6}=-4.167 \ \text{cm}$
$\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}$
$\frac{1}{\text{u}}+\frac{1}{\text{v}}=\frac{1}{\text{f}}$
$=\frac{1}{15}+\frac{1}{12}=\frac{4+5}{60}=\frac{9}{60}$
$\therefore \ \text{v}=\frac{60}{9}=6.7 \ \text{cm}$
Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror.
The image size is given by the magnification formula:
$\text{m}=\frac{\text{h}_2}{\text{h}_1}=-\frac{\text{v}}{\text{u}}$
$\therefore \ \text{h}_2=-\frac{\text{v}}{\text{u}}\times\text{h}_1$
$=\frac{-6.7}{-12}\times4.5=+2.5 \ \text{cm}$
Hence, magnification of the image, $\text{m}=\frac{\text{h}_2}{\text{h}_1}=\frac{2.5}{4.5}=0.56$
The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual, and diminished.
If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually.

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Question 85 Marks

A card sheet divided into squares each of size $1\ mm^2$ is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye.

What is the magnification produced by the lens? How much is the area of each square in the virtual image?
What is the angular magnification (magnifying power) of the lens?
Is the magnification in (a) equal to the magnifying power in (b)? Explain.

Answer

Given,

Object distance, u = -9 cm
Focal length of converging lens, f = 10 cm
Now, Using the formula,
$\Rightarrow \ \frac{1}{\text{v}}=\frac{1}{\text{f}}+\frac{1}{\text{u}}$
$\Rightarrow \ \frac{1}{\text{v}}=\frac{1}{10}+\frac{1}{-9}$
$=\frac{-9+10}{-90}$
$=\frac{1}{-90}$
i.e., v = -90 cm
Magnification, $\text{m}=\frac{\text{v}}{\text{u}}=\frac{-90}{-9}=10$
Therefore,
Area of the each square in the virtual image $= (10 \times 10 \times 1) mm^2 = 100 mm^2 = 1 cm^2.$

Given,

Least distance of distinct vision, D = -25 cm
Object distance, u = -9 cm
Therefore,
Angular magnification $=\frac{\text{D}}{\text{u}}=\frac{-25}{-9}=2.78$

No. Magnification of an image by a lens and angular magnification (or magnifying power) of an optical instrument are different from each other.

Angular magnification is the ratio of the angular size of the image ( even if the image is magnified) to the angular size of the object if placed at the near point (25 cm).
Thus, magnification magnitude $=|\frac{\text{v}}{\text{u}}|$ and,
Magnifying power is $\frac{25}{|\text{u}|}=\frac{\text{D}}{\text{u}}$.
Only when the image is located at the least distance of distinct vision, we can have both the quantities equal.

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Question 95 Marks

At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

Answer

The refracted ray in the prism is incident on the second face at critical angle ic.

Given, angle of prism, A = 60°
Refracting index of the material = 1.524
Now,$ 60^\circ + 90^\circ - r + 90^\circ - i_c=_180^\circ$
and $\text{r}_1+\text{r}_2=\angle\text{A}$
$\Rightarrow \ \text{r}=60^\circ-\text{i}_\text{c}$
$\sin\text{i}_\text{c}=\frac{1}{\mu}=\frac{1}{1.524}$
$\Rightarrow \ \text{i}_\text{c}=\sin^{-1}\Big(\frac{1}{1.524}\Big)$
$\Rightarrow \ \text{i}_\text{c}\simeq41^\circ$
$\therefore \ \text{r}=60^\circ-41^\circ=19^\circ$
Using Snell's law,
$\sin\text{i}=\sin19^\circ\times1.524$
= 0.4962
$\text{i}=\sin^{-1}(0.4962)$
= 29.75°.

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Question 105 Marks

A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.

What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?
What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?

Answer

For the closest distance,

Image distance, v = -25 cm

Focal length of thin convex lens, f = 5 cm

Using the formula,

$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\therefore \ \frac{1}{\text{u}}=\frac{1}{\text{v}}-\frac{1}{\text{f}}$

$\frac{1}{-25}-\frac{1}{\text{u}}=\frac{1}{5}$

$-\frac{1}{\text{u}}=\frac{1}{5}+\frac{1}{25}$

$-\frac{1}{\text{u}}=\frac{5+1}{25}$

$-\frac{1}{\text{u}}=\frac{6}{25}$

$-\text{u}=\frac{25}{6}$

$\therefore\ \text{u}=-\frac{25}{6}=-4.2 \ \text{cm}$

l.e.. 4.2 cm is the closest distance at which the man can read the book.

For the farthest image,

Image distance, v = $\infty$

Focal length, f = 5 cm

Using the formula,

$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\frac{1}{\text{u}}=\frac{1}{\text{v}}-\frac{1}{\text{f}}$

$=\frac{1}{\infty}-\frac{1}{5}$

$=-\frac{1}{5}$

$\text{u} = -5 \text{cm}$

which is the object distance

This is the farthest distance at which the man can read the book.

Normal near point of human eye = 25 cm

Focal length , f = 5 cm

Maximum angular magnification is

$\frac{\text{D}}{\text{U}_\text{min}}=\frac{25}{\frac{25}{6}}=6$

Minimum angular magnification is

$\frac{\text{D}}{\text{U}_\text{max}}=\frac{25}{5}=5.$

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Question 115 Marks

An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.

Answer

For convex lens, Object distance from the lens, u = -40cm Focal length, f = 30 cm Object size, 0 = 1.5 cm Using lens formula $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$ We get, $\frac{1}{\text{v}}-\frac{1}{-40}=\frac{1}{30}$ $\frac{1}{\text{v}}=\frac{1}{30}-\frac{1}{40}=\frac{1}{120}$i.e., v= 120 cm (for real object)
From relation, Magnification, $\text{m}=-\frac{\text{v}}{\text{u}}$, we get $\text{m}=-\frac{120}{-40}=+3$ The image formed by the convex lens becomes object for concave lens at a distance of (120 - 8) = 112 cm on the other side. For concave lens, Focal length, f = - 20 cm Object distance, u = +112 cm (on the other side) Image distance, v = ? Using lens formula, we get $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$ Now, $\frac{1}{\text{v}}-\frac{1}{112}=\frac{1}{-20}$ $\Rightarrow \ \frac{1}{\text{v}}=-\frac{1}{20}+\frac{1}{112}$ $=-\frac{23}{560}$ $\text{v}=-\frac{560}{23} \ \text{cm}$ (for virtual object) Using the formula of magnification m = $\frac{\text{v}}{\text{u}}$, we get $\text{m}=-\frac{\frac{560}{23}}{-112}$ $=-\frac{560}{23}\times\frac{1}{-112}$ $=-\frac{5}{23}$ Net magnification, $\text{m}=3\times\Big(\frac{-5}{23}\Big)$ $=-\frac{15}{23}$ = 0. 652 ( negative due to virtual image) and as, $\text{m}=\frac{\text{I}}{0}$ I = m × 0 = 0.652 × 1.5 = 0.98 cm (size of final image).

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Question 125 Marks

A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Answer

Focal length of the objective lens, $f_1 = 2.0$ cm
Focal length of the eyepiece, $f_2 = 6.25$ cm
Distance between the objective lens and the eyepiece, d = 15 cm

Least distance of distinct vision, d' = 25 cm

$\therefore$ Image distance for the eyepiece, $v_2 = -25$ cm
Object distance for the eyepiece $= u_2$
According to the lens formula, we have the relation:
$\frac{1}{\text{v}_2}-\frac{1}{\text{u}_2}=\frac{1}{\text{f}_2}$
$\frac{1}{\text{u}_2}=\frac{1}{\text{v}_2}-\frac{1}{\text{f}_2}$
$=\frac{1}{-25}-\frac{1}{6.25}=\frac{-1-4}{25}=\frac{-5}{25}$
$\therefore \ \text{u}_2=-5 \ \text{cm}$
Image distance for the objective lens, $\text{v}_1=\text{d}+\text{u}_2=15-5=10 \ \text{cm}$
Object distance for the objective lens $= u_2$
According to the lens formula, we have the relation:
$\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}_1}$
$\frac{1}{\text{u}_1}=\frac{1}{\text{v}_1}-\frac{1}{\text{f}_1}$
$=\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=\frac{-4}{10}$
$\therefore \ \text{u}_1=-2.5\ \text{cm}$
Magnitude of the object distance, $|u_1| = 2.5$ cm
The magnifying power of a compound microscope is given by the relation:
$\text{m}=\frac{\text{v}_1}{|\text{u}_1|}\Big(1+\frac{\text{d}'}{\text{f}_2}\Big)$
$=\frac{10}{2.5}\Big(1+\frac{25}{6.25}\Big)=4(1+4)=20$
Hence, the magnifying power of the microscope is 20.

The final image is formed at infinity.

$\therefore$ Image distance for the eyepiece, $v_2 = \infty$
Object distance for the eyepiece $= u_2$
According to the lens formula, we have the relation:
$\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}_1}$
$\frac{1}{\text{v}_2}-\frac{1}{\text{u}_2}=\frac{1}{\text{f}_2}$
$\frac{1}{\infty}-\frac{1}{\text{u}_2}=\frac{1}{6.25}$
$\therefore \ \text{u}_2=-6.25\ \text{cm}$
Image distance for the objective lens, $\text{v}_1=\text{d}+\text{u}_2=15-6.25=8.75 \ \text{cm}$
Object distance for the objective lens $= u_1$
According to the lens formula, we have the relation:
$\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}_1}$
$\frac{1}{\text{u}_1}=\frac{1}{\text{v}_1}-\frac{1}{\text{f}_1}$
$=\frac{1}{8.75}-\frac{1}{2.0}=\frac{2-8.75}{17.5}$
$\therefore \ \text{u}_1=-\frac{17.5}{6.75}=-2.59\ \text{cm}$
Magnitude of the object distance, $u_1 = 2.59$ cm
The magnifying power of a compound microscope is given by the relation:
$\text{m}=\frac{\text{v}_1}{|\text{u}_1|}\Bigg(\frac{\text{d}'}{|\text{u}_2|}\Bigg)$
$=\frac{8.75}{2.59}\times\frac{25}{6.25}=13.51$
Hence, the magnifying power of the microscope is 13.51.

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Question 135 Marks

A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Answer

Focal length of the objective lens, $f_1 = 2.0$ cm
Focal length of the eyepiece, $f_2 = 6.25$ cm
Distance between the objective lens and the eyepiece, d = 15 cm

Least distance of distinct vision, $d' = 25$ cm

$\therefore$ Image distance for the eyepiece, $v_2 = -25$ cm
Object distance for the eyepiece $= u_2$
According to the lens formula, we have the relation:
$\frac{1}{\text{v}_2}-\frac{1}{\text{u}_2}=\frac{1}{\text{f}_2}$
$\frac{1}{\text{u}_2}=\frac{1}{\text{v}_2}-\frac{1}{\text{f}_2}$
$=\frac{1}{-25}-\frac{1}{6.25}=\frac{-1-4}{25}=\frac{-5}{25}$
$\therefore \ \text{u}_2=-5 \ \text{cm}$
Image distance for the objective lens, $\text{v}_1=\text{d}+\text{u}_2=15-5=10 \ \text{cm}$
Object distance for the objective lens $= u_2$
According to the lens formula, we have the relation:
$\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}_1}$
$\frac{1}{\text{u}_1}=\frac{1}{\text{v}_1}-\frac{1}{\text{f}_1}$
$=\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=\frac{-4}{10}$
$\therefore \ \text{u}_1=-2.5\ \text{cm}$
Magnitude of the object distance, $|u_1| = 2.5$ cm
The magnifying power of a compound microscope is given by the relation:
$\text{m}=\frac{\text{v}_1}{|\text{u}_1|}\Big(1+\frac{\text{d}'}{\text{f}_2}\Big)$
$=\frac{10}{2.5}\Big(1+\frac{25}{6.25}\Big)=4(1+4)=20$
Hence, the magnifying power of the microscope is 20.

The final image is formed at infinity.

$\therefore$ Image distance for the eyepiece, $v_2 = \infty $
Object distance for the eyepiece $= u_2$
According to the lens formula, we have the relation:
$\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}_1}$
$\frac{1}{\text{v}_2}-\frac{1}{\text{u}_2}=\frac{1}{\text{f}_2}$
$\frac{1}{\infty}-\frac{1}{\text{u}_2}=\frac{1}{6.25}$
$\therefore \ \text{u}_2=-6.25\ \text{cm}$
Image distance for the objective lens, $\text{v}_1=\text{d}+\text{u}_2=15-6.25=8.75 \ \text{cm}$
Object distance for the objective lens $= u_1$
According to the lens formula, we have the relation:
$\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}_1}$
$\frac{1}{\text{u}_1}=\frac{1}{\text{v}_1}-\frac{1}{\text{f}_1}$
$=\frac{1}{8.75}-\frac{1}{2.0}=\frac{2-8.75}{17.5}$
$\therefore \ \text{u}_1=-\frac{17.5}{6.75}=-2.59\ \text{cm}$
Magnitude of the object distance, $u_1 = 2.59$ cm
The magnifying power of a compound microscope is given by the relation:
$\text{m}=\frac{\text{v}_1}{|\text{u}_1|}\Bigg(\frac{\text{d}'}{|\text{u}_2|}\Bigg)$
$=\frac{8.75}{2.59}\times\frac{25}{6.25}=13.51$
Hence, the magnifying power of the microscope is 13.51.

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Question 145 Marks

equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?

Answer

Focal length of the convex lens, $f_1 = 30$ cm
The liquid acts as a mirror. Focal length of the liquid $= f_2$
Focal length of the system (convex lens + liquid), $f = 45$ cm
For a pair of optical systems placed in contact, the equivalent focal length is given as:
$\frac{1}{\text{f}}=\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}$
$\frac{1}{\text{f}_2}=\frac{1}{\text{f}}-\frac{1}{\text{f}_1}$
$=\frac{1}{45}-\frac{1}{30}=-\frac{1}{90}$
$\therefore \ \text{f}_2=-90 \ \text{cm}$
Let the refractive index of the lens be $µ_1$ and the radius of curvature of one surface be R.
Hence, the radius of curvature of the other surface is - R.
R can be obtained using the relation:
$\frac{1}{\text{f}_1}=(\mu-1)\Big(\frac{1}{\text{R}}+\frac{1}{\text{R}}\Big)$
$\frac{1}{30}=(1.5-1)\Big(\frac{2}{\text{R}}\Big)$
$\therefore \ \text{R}=30\times(0.5)\times2=30\text{cm}$
Let $µ_2$ be the refractive index of the liquid.
Radius of curvature of the liquid on the side of the plane mirror = ∞
Radius of curvature of the liquid on the side of the lens, R= - 30 cm
The value of $µ_2$ can be calculated using the relation:
$\frac{1}{\text{f}_2}=(\mu_2-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$
$-\frac{1}{90}=(\mu_2-1)\Big(\frac{1}{-30}-\frac{1}{\infty}\Big)$
$\frac{1}{90}=(\mu_2-1)\Big(\frac{1}{30}\Big)$
$\mu_2-1=\frac{30}{90}$
$\mu_2=\frac{1}{3}+1=\frac{4}{3}$
$=1.33$

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Question 155 Marks

A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.

Answer

Focal length of the objective lens, $f_0 = 8 mm = 0.8 cm$
Focal length of the eyepiece, $f_e = 2.5$ cm
Object distance for the objective lens, $u_0 = - 9.0 mm = - 0.9$ cm
Least distance of distant vision, $d = 25$ cm
Image distance for the eyepiece, $v_e = - d= - 25$ cm
Object distance for the eyepiece $= u_e$ Using the lens formula,
we can obtain the value of $u_e$
as: $\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{f}_\text{e}}$ $\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{f}_\text{e}}$
$=\frac{1}{-25}-\frac{1}{2.5}=\frac{-1-10}{25}=\frac{-11}{25}$
$\therefore \ \text{u}_\text{e}=-\frac{25}{11}=-2.27 \ \text{cm}$
We can also obtain the value of the image distance for the objective lens ( ) 0 v using the lens formula.
$\frac{1}{\text{v}_0}-\frac{1}{\text{u}_0}=\frac{1}{\text{f}_0}$
$\frac{1}{\text{v}_0}=\frac{1}{\text{f}_0}+\frac{1}{\text{u}_0}$
$=\frac{1}{0.8}-\frac{1}{0.9}=\frac{0.9-0.8}{0.72}=\frac{0.1}{0.72}$
$\therefore \ \text{v}_0=7.2 \ \text{cm}$
The distance between the objective lens and the eyepiece $= |u_e| + v_0 = 2.27 + 7.2 = 9.47\ cm$
The magnifying power of the microscope is calculated as:
According to the lens formula, we have the relation:
$\frac{\text{v}_0}{|\text{u}_0|}\Big(1+\frac{\text{d}}{\text{f}_\text{e}}\Big)$ $=\frac{7.2}{0.9} \Big(1+\frac{25}{2.5}\Big)=8(1+10)=88$
Hence, the magnifying power of the microscope is 88.

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Question 165 Marks

A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source).

Answer

Actual depth of the bulb in water,$ d_1 = 80 cm = 0.8 m$
Refractive index of water, $µ = 1.33$
The given situation is shown in the following figure:
Where,
i = Angle of incidence
r = Angle of refraction = 90°
Since the bulb is a point source, the emergent light can be considered as a circle of radius,
$\text{R}=\frac{\text{AC}}{2}=\text{AO}=\text{OB}$
Using Snell' law, we can write the relation for the refractive index of water as:
$\mu=\frac{\sin\text{r}}{\sin\text{i}}$
$1.33=\frac{\sin90^\circ}{\sin\text{i}}$
$\therefore \ \text{i}=\sin^{-1}\Big(\frac{1}{1.33}\Big)=48.75^\circ$
Using the given figure, we have the relation:
$\tan\text{i}=\frac{\text{OC}}{\text{OB}}=\frac{\text{R}}{\text{d}_1}$
$\therefore \ \text{R}=\tan48.75^\circ\times0.8=0.91\text{m}$
$\therefore \ $Area of the surface of water $=\pi\text{R}^2=\pi(0.91)^2=2.61\text{m}^2$
Hence, the area of the surface of water through which the light from the bulb can emerge is approximately $2.61 m^2.$

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Question 175 Marks

(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45º with the normal to a water - glass interface.

Answer

As per the given figure, for the glass - air interface: Angle of incidence, i = 60° Angle of refraction, r= 35° The relative refractive index of glass with respect to air is given by Snell's law as: $=\text{Mh}_2=6\times4.7=28.2 \ \text{cm}$ $\mu^\circ_\text{g}=\frac{\sin\text{i}}{\sin\text{r}}$ $\frac{\sin 60^\circ}{\sin35^\circ}=\frac{0.8660}{0.5736}=1.51\dots(1)$ As per the given figure, for the air - water interface: Angle of incidence, i = 60° Angle of refraction, r= 47° The relative refractive index of water with respect to air is given by Snell's law as: $\mu^\circ_\text{g}=\frac{\sin\text{i}}{\sin\text{r}}$ $\frac{\sin 60^\circ}{\sin47^\circ}=\frac{0.8660}{0.5736}=1.184\dots(2)$ Using (1) and (2), the relative refractive index of glass with respect to water can be obtained as: $\mu^\text{w}_\text{g}=\frac{\mu^\text{a}_\text{g}}{\mu^\text{a}_\text{w}}$ $=\frac{1.51}{1.184}=1.275$ The following figure shows the situation involving the glass - water interface.

Angle of incidence, i= 45° Angle of refraction = r From Snell's law, rcan be calculated as: $\frac{\sin\text{i}}{\sin\text{r}}=\mu^\text{w}_\text{g}$ $\frac{\sin45^\circ}{\sin\text{r}}=1.275$ $\sin\text{r}=\frac{\frac{1}{\sqrt2}}{1.275}=0.5546$ $\therefore \ \text{r}=\sin^{-1}(0.5546)=38.67^\circ$ Hence, the angle of refraction at the water - glass interface is 38.68°.

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Question 185 Marks

Derive the mathematical relation between refractive indices $n_1$ and $n_2 $ of two radii and radius of curvature R for refraction at a convex spherical surface. Consider the object to be a point since lying on the principle axis in rarer medium of refractive index $n_1$ and a real image formed in the denser medium of refractive index $n_{2}$.Hence, derive lens maker's formula.
Light from a point source in air falls on a convex spherical glass surface of refractive index 1.5 and radius of curvature 20 cm. The distance of light source from the glass surface is 100 cm. At what position is the image formed?

Answer

Ray diagram showing real image formation as per prescription,
$\theta_{1} =\alpha + \beta$
$\theta_{2} = \beta - \gamma$
$\therefore \gamma = \beta - \theta $
For paraxial rays $\theta_{1}$ and $\theta_{2}$ are small Therefore,$ n_2 \sin \theta_{2} = n_1 \sin \theta_{2}$ (Snells law)
Reduces to,
At N $\frac{\sin\text{i}}{\sin\text{r}}\sim\frac{\text{i}}{\text{r}} = \frac{\text{n}_{2}}{\text{n}_{1}}$
$\therefore\text{n}_{1} =\text{rXn}_{2}$
$( \alpha +\beta ) \text{n}_{1} = (\beta - \theta ) \text{n}_{2}$
$\text{n}_{1}\big(\frac{\text{NM}}{\text{OM}} + \frac{\text{NM}}{\text{MC}}\big) = \big(\frac{\text{NM}}{\text{MC}} - \frac{\text{NM}}{\text{MI}}\big)\text{n}_{2}$
$\text{n}_{1}\big(\frac{1}{\text{ - u}} +\frac{1}{\text{ + R }}\big) = \big(\frac{1}{\text{ + R}} - \frac{1}{\text{u}}\big)\text{n}_{2}$
$\frac{\text{n}_{2}}{\text{v}'} - \frac{\text{n}_{1}}{\text{u}} = \frac{(\text{n}_{2} - \text{n}_{1})}{\text{R}_{1}}$
Appying above relations to refraction through a lens:

For surface 1
$\frac{\text{n}_{2} - \text{n}_{1}}{\text{R}_{1}} = \frac{\text{n}_{1}}{\text{v}'} - \frac{\text{n}_{1}}{\text{u}}$ . ... . ... .. (i)
For surface 2
$\frac{\text{n}_{1} - \text{n}_{2}}{\text{R}_{2}} = \frac{\text{n}_{1}}{\text{v}} - \frac{\text{n}_{2}}{\text{v}'}$ . . .. . . . . (ii)
Adding eqn. (i) and (ii)
$(\text{n}_{2} - \text{n}_{1}) \big(\frac{1}{\text{R}_{1}} -\frac{1}{\text{R}_{2}}\big) = \text{n}_{1}\big(\frac{1}{\text{v}} - \frac{1}{\text{u}}\big)$
For $\text{u}=\propto \text{v}=\text{f}$
$\therefore \frac{\text{n}_{1}}{\text{f}} = (\text{n}_{2} - \text{n}_{1})\big(\frac{1}{\text{R}_{1}} - \frac{1}{\text{R}_{2}}\big)$
$\Rightarrow\frac{1}{\text{f}} = \bigg(\frac{\text{n}_{2}}{\text{n}_{1}} - 1 \bigg)\bigg(\frac{1}{\text{R}_{1}} - \frac{1}{\text{R}_{2}}\bigg)$

$R = 20$ cm $n_2 = 1.5$ $n_1 = 1$ $u = - 100$ cm

$\frac{\text{n}_{2}}{\text{v}} = \frac{(\text{n}_{2} - \text{n}_{1})}{\text{R}} + \frac{\text{n}_{1}}{\text{u}}$
$ = \frac{0.5}{20\text{cm}} - \frac{1}{100\text{cm}}$
$ =\frac{1.5}{100}\text{cm}$
$\Rightarrow\text{V} = 100 \text{ cm}$ a real image on the other side, 100 cm away from the surface.

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Question 195 Marks

Draw a labelled ray diagram to obtain the real image formed by an astronomical telescope in normal adjustment position. Define its magnifying power.
You are given three lenses of power 0.5 D, 4 D and 10 D to design a telescope:

Which lenses should he used as objective and eyepiece? Justify your answer.
Why is the aperture of the objective preferred to be large?

Answer

Definition - It is the ratio of the angle subtended at the eye, by the final image, to the angle which the object subtends at the lens, or the eye.

Objective=.5D

Eye lens = 10D This choice would give higher magnification as.
$\text{M} = \frac{\text{f}_{o}}{\text{f}_{e}} = \frac{\text{p}_{e}}{\text{p}_{o}}$

High resolving power/Brighter image/lower limit of resolution.

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Question 205 Marks

Draw a labelled ray diagram to obtain the real image formed by an astronomical telescope in normal adjustment position. Defineits magnifying power.
You are given three lenses of power 0.5 D, 4 D and 10 D to design a telescope.

Which lenses should he used as objective and eyepiece? Justify your answer.
Why is the aperture of the objective preferred to be large?

Answer

Definition - It is the ratio of the angle subtended at the eye, by the final image, to the angle which the object subtends at the lens, or the eye.

Objective=.5D

Eye lens = 10D This choice would give higher magnification as.
$\text{M} = \frac{\text{f}_{o}}{\text{f}_{e}} = \frac{\text{p}_{e}}{\text{p}_{o}}$

High resolving power/Brighter image/lower limit of resolution.

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Question 215 Marks

Draw a labelled ray diagram showing the formation of image by a compound microscope in normal adjustment. Derive the expression for its magnifying power.
How does the resolving power of a microscope change when.

The diameter of the objective lens is decreased,
The wavelength of the incident light is increased.

Justify your answer in each case.

Answer

Derivation:
- Magnification due to objective $\text{M}_{\circ} = \frac{\text{L}}{\text{f}_{\circ}}$
- Magnification due to eyelens $\text{M}_{e} = \frac{\text{D}}{\text{f}_{e}}$
- Total magnification m $ = \text{m}_{\circ}\text{m}_{e}$
$\text{m}_{\circ} = \frac{\text{L}}{\text{f}_{\circ}}.\frac{\text{D}}{\text{f}_{e}}$

The resolving power of microscope.

Will decrease with decrease of the diameter of objective lens as resolving power is directly proportional to the diameter.
Will decrease with increase of the wavelength of the incident light as resolving power is inversely proportional to the wave length.

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Question 225 Marks

Draw a ray diagram to show the image formation by a combination of two thin convex lenses in contact. Obtain the expression for the power of this combination in terms of the focal lengths of the lenses.
A ray of light passing from air through an equilateral glass prism undergoes minimum deviation when the angle of incidence is $\frac{3}{4}^{th}$ of the angle of prism. Calculate the speed of light in the prism.

Answer

Two thin lenses, of focal length $f_1$ and $f_2$ are kept in contact. Let O be the position of object and let u be the object distance. The distance of the image (which is at $I_1$), for the first lens is $v_1$. This image serves as object for the second lens.
Let the final image be at I. We then have
$\frac{1}{f_1}=\frac{1}{v_1}-\frac{1}{u}$
$\frac{1}{f_2}=\frac{1}{v}-\frac{1}{v_1}$
Adding, we get
$\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\therefore\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$
$\therefore P=P_1+P_2$
At minimum deviation
$r=\frac{A}{2}=30^\circ$
We are given that
$i=\frac{3}{4}A=45^\circ$
$\therefore\text{ }\mu=\frac{\sin 45^\circ}{\sin 30^\circ}=\sqrt{2}$
$\therefore$ Speed of light in the prism $=\frac{c}{\sqrt{2}}$
$(\cong2.1\times10^8 \text{ms}^{-1})$

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Question 235 Marks

One day Chetan’s mother developed a severe stomach ache all of a sudden. She was rushed to the doctor who suggested for an immediate endoscopy test and gave an estimate of expenditure for the same. Chetan immediately contacted his class teacher and shared the information with her. The class teacher arranged for the money and rushed to the hospital. On realising that Chetan belonged to a below average income group family, even the doctor offered concession for the test fee. The test was conducted successfully.
Answer the following questions based on the above information:

Which principle in optics is made use of in endoscopy?
Briefly explain the values reflected in the action taken by the teacher.
In what way do you appreciate the response of the doctor on the given situation?

Answer

Total internal reflection: If a light ray enters at one end of an optic fibre coated with a material of low refractive index, it refracted and strikes the walls at angle greater than critical angle. Thus light rays shows multiple reflections, without being absorbed at the side walls.

The teacher knows that Chetan belongs to a below average income group family, so he/she immediatelly arranged the money required to be paid as test fee. His/her caring and helping attitude towards the others resulted in timely help to Chetan’s mother. Such helping attitude on the part of the person living in the society make it a better society to live in.
Seeing the situation of Chetan’s family and helping attitude of class teacher, doctor took the sympathetic view of the situation, and give the reduction in fee, which is highly appreciable. Such professional ethics of doctor in the society would be an immense help to the person’s belonging to below average income groups.

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Question 245 Marks

Draw a labelled ray diagram to show the formation of image in an astronomical telescope for a distant object.
Write three distinct advantages of a reflecting type telescope over a refracting type telescope.

A convex lens of focal length 10 cm is placed coaxially 5 cm away from a concave lens of focal length 10 cm. If an object is placed 30 cm in front of the convex lens, find the position of the final image formed by the combined system.

Answer

Alternate Answer

Advantages:
1. Less chromatic abberation.
2. High resolving power.
3. Large gathering power.
4. Less spherical abberation.

The position of the image, formed by the convex lens, is given by
$\frac{1}{\text{v}'} - \frac{1}{30} = \frac{1}{10}$
$\therefore\text{v}' = 15\text{cm}$
$\therefore\text{For the concave lens, u =+(15 − 5 )cm}$
$\text{and }\text{f} = -10\text{cm}$
$\text{Hence }\frac{1}{\text{v}} -\frac{1}{10} = - \frac{1}{10}$
$\therefore\frac{1}{\text{v}} = 0 \text{ or } \text{v}\rightarrow\infty.$

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Question 255 Marks

With the help of a suitable ray diagram, derive the mirror formula for a concave mirror.
The near point of a hypermetropic person is 50 cm from the eye. What is the power of the lens required to enable the person to read clearly a book held at 25 cm from the eye?

Answer

Derivation

Triangle A' B'F is similar to $\Delta$MPF.

Therefore $\frac{\text{B}'\text{A}'}{\text{PM}} =\frac{\text{B}'\text{F}}{\text{FP}}$

From Fig. PM = BA

$\text{Right angled}\Delta^\text{S}\text{A}'\text{B}'\text{P}\text{ and ABP are similar.}$

$\frac{\text{B}'\text{A}'}{\text{BA}} = \frac{\text{B}'\text{P}}{\text{BP}}$

On comparing $\frac{\text{B}'\text{F}}{\text{FP}} = \frac{\text{B}'\text{P}}{\text{BP}}$

$\frac{-\text{v} + \text{f}}{-\text{f}} =\frac{-\text{v}}{\text{-u}}\Rightarrow$

$\frac{1}{\text{f}} =\frac{1}{\text{v}} + \frac{1}{\text{u}}$

$\frac{1}{\text{v}} - \frac{1}{\text{u}} = \frac{1}{\text{f}}$

$\frac{1}{50} + \frac{1}{25} = \frac{1}{\text{f}}$

$\therefore\text{f} = 50\text{ cm} = 0.5\text{m}$

$\text{Power =}\frac{1}{\text{f}} = \frac{1}{0.5}\text{D}$

$ = 2\text{D}.$

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Question 265 Marks

Derive the lens formula, $\frac{1}{\text{f}} = \frac{1}{\text{v}} - \frac{1}{\text{u}}$ for a concave lens, using the necessary ray diagram.
Two lenses of powers 10 D and – 5 D are placed in contact.

Calculate the power of the new lens.
Where should an object be held from the lens, so as to obtain a virtual image of magnification 2?

Answer

Diagram:

Derivation: $\frac{1}{\text{v}} -\frac{1}{\text{u}} = \frac{1}{\text{f}}$

$\text{Power} = \text{P}_{1} + \text{P}_{2}$

$ = 10 - 5 = 5\text{D}$

f = 20 cm

$\text{m} = \frac{\text{v}}{\text{u}} = 2$

$\frac{1}{\text{f}} =\frac{1}{\text{v}} - \frac{1}{\text{u}}$

$\text{u} = -10\text{cm}.$

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Question 275 Marks

A point object is placed on the principal axis of a convex spherical surface of radius of curvature R, which separates the two media of refractive indices $n_1$ and $n_2(n_2 > n_1)$. Draw the ray diagram and deduce the relation between the object distance (u), image distance (v) and the radius of curvature (R) for refraction to take place at the convex spherical surface from rarer to denser medium.
A converging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1.6. If it is immersed in a liquid of refractive index 1.3, find its new focal length.

Answer

For small angles
$\tan\angle NOM=\frac{MN}{OM} :\tan\angle NCM=\frac{MN}{NC}$
and $\tan\angle NIM=\frac{MN}{MI}$
For Δ???, i is exterior angle, therefore
$\text{i}=\angle NOM+\angle NCM=\frac{MN}{OM}+\frac{MN}{MC}$
Similarly $\text{r}=\frac{MN}{MC}-\frac{MN}{MI}$
For small angles Snells law can be written as
$n_1\text{i}=n_2r$
$\therefore\text{ }\frac{n_1}{OM}+\frac{n_2}{MI}=\frac{n_2-n_1}{MC}$
$\therefore\text{ }\text{OM}=\text{-u},\text{ }\text{MI}=+\text{v},\text{ }\text{MC}=+\text{R}$ (using sign conversion)
$\therefore\text{ }\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$

Lens Maker’s formula is

$\frac{1}{f_a}=\big(\frac{n_2-1}{n_1}\big)(\frac{1}{R_1}-\frac{1}{R_2})$
$\therefore\frac{1}{20}=(1.6-1)(\frac{1}{R_1}-\frac{1}{R_2})$
$\therefore\big(\frac{1}{R_1}-\frac{1}{R_2}\big)=\frac{1}{20\times0.6}=\frac{1}{12}$
Let f be the focal length of the lens in water
$\therefore\frac{1}{f'}=\frac{1.6-1.3}{1.3}\big(\frac{1}{R_1}-\frac{1}{R_2}\big)=\frac{0.3}{12\times1.3}$
$\frac{1}{f'}=\frac{120\times1.3}{3}=52\text{ }cm$

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Question 285 Marks

Draw the ray diagram showing refraction of light through a glass prism and hence obtain the relation between the refractive index $\mu$ of the prism, angle of prism and angle of minimum deviation.
Determine the value of the angle of incidence for a ray of light travelling from a medium of refractive index $\mu_1=\sqrt{2}$ into the medium of refractive index $\mu_2=1$, so that it just grazes along the surface of separation.

Answer

From fig $\angle A+\angle QNR=180^\circ\text{ }\text{ }\text{ }\text{ }\text{ }\dots(1)$
From triangle $\triangle QNR \text{ }\text{ }r_1+r_2+\angle QNR=180^\circ\text{ }\text{ }\text{ }\text{ }\dots(2)$
Hence from equ (1) & (2)
$\therefore \angle A= r_1+r_2$
The angle of deviation
$\delta=(i-r_1)+(e-r_2)=\text{i + e - A}$
At minimum deviation i = e and $?_1 = ?_2$
$\therefore r=\frac{A}{2}$
And $\text{i}=\frac{A+\delta m}{2}$
Hence refractive index
$\mu=\frac{\sin i}{\sin r}=\frac{\sin\big(\frac{A+\delta}{2}\big)}{\sin\frac{A}{2}}$

From Snell’s law $\mu_1 \sin i=\mu_2\sin r$

Given $\mu_1=\sqrt{2}, \mu_2=1$ and $r = 90^0$ (just grazing)
$\therefore\sqrt{2}\text{ }\sin \text{i}=1\sin90^\circ\Rightarrow \sin i\frac{1}{\sqrt{2}}$
$or \text{ }i=45^\circ$

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Question 295 Marks

Mrs. Rashmi Singh broke her reading glasses. When she went to the shopkeeper to order new spects, he suggested that she should get spectacles with plastic lenses instead of glass lenses. On getting the new spectacles, she found that the new ones were thicker than the earlier ones. She asked this question to the shopkeeper but he could not offer satisfactory explanation for this. At home, Mrs. Singh raised the same question to her daughter Anuja who explained why plastic lenses were thicker.

Write two qualities displayed each by Anuja and her mother.
How do you explain this fact using lens maker's formula?

Answer

Anuja: Scientific temperament, cooperative, knowledgeable.

Mother: Inquisitive, scientific temper/keen to learn/has no airs.

$\frac{1}{f}=\Big(\frac{\mu_2}{\mu_1}-1\Big)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)$

As the refractive index of plastic material is less than that of glass material therefore, for the same power $\big(=\frac{1}{f}\big)$, the radius of currature of plastic material is small.

Therefore plastic lens is thicker.

Alternatively, If student just writes that plastic has a different refractive index than glass, award one mark for this part.

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Question 305 Marks

A point object 'O' is kept in a medium of refractive index $n_1$ in front of a convex spherical surface of radius of curvature R which separates the second medium of refractive index $n_2$ from the first one, as shown in the figure.

Draw the ray diagram showing the image formation and deduce the relationship between the object distance and the image distance in terms of $n_1, n_2$ and R.
When the image formed above acts as a virtual object for concave spherical, surface separating the medium $n_2$ from $n_1 (n_2> n_1)$, draw this ray diagram and write the similar (similar to (a) relation. Hence obtain the expression for the lens maker's formula.

Answer

For small angles
$\angle\text{N0M}\cong\tan\angle \text{N0M} = \frac{\text{MN}}{\text{OM}}$
$\angle\text{NCM}\cong\tan\angle \text{NCM} = \frac{\text{MN}}{\text{MC}}$
$\angle\text{NIM}\cong\tan\angle \text{NIM} = \frac{\text{MN}}{\text{MI}}$
In $\Delta\text{NOC} ,$ $\angle \text{i} = \angle\text{i}\angle\text{NOM} + \angle\text{NCM}$
$\therefore\angle\text{i} = \frac{\text{MN}}{\text{OM}} + \frac{\text{MN}}{\text{MC}}$
$\angle\text{r} = \angle \text{NCM} -\angle \text{NIM}$
$=\frac{\text{MN}}{\text{MC}} -\frac{\text{MN}}{\text{MI}}$
Using Snell's Law
$n_1 \sin\ i. = n_2 \sin\ r$
For small angles $n_1\ i = n_2\ r$
Substituting for i and r, we get
$\frac{\text{n}_{1}}{\text{OM}} +\frac{\text{n}_{2}}{\text{MI}} = \frac{\text{n}_{2} -\text{n}_{1}}{\text{MC}}$
Here, $\text{OM} =\text{-u}, \text{MI} = + \text{v},\text{MC} = + \text{R}$
Substituting these, we get
$\Rightarrow\frac{\text{n}_{2}}{\text{v}} - \frac{\text{n}_{1}}{\text{u}} = \frac{\text{n}_{2} -\text{n}_{1}}{\text{R}}$

(Alternatively accept this Ray diagram)

Similarly relation for the surface ADC.
$\frac{-\text{n}}{\text{DI}{1}} +\frac{\text{n}_{1}}{\text{DI}} = \frac{\text{n}_{2} - \text{n}_{1}}{\text{DC}_{2}}$ ................................(i)
Refraction at the first surface ABC of the lens.
$\frac{\text{n}_{1}}{\text{OB}} +\frac{\text{n}_{2}}{\text{BI}_{1}} = \frac{\text{n}_{2} -\text{n}_{2}}{\text{BC}_{1}}$..............................(ii)
Adding (i)and (ii), and taking $BI_1\cong_DI_1, $ we get
$\frac{\text{n}_{1}}{\text{OB}} +\frac{\text{n}_{1}}{\text{DI}} = (\text{n}_{2} -\text{n}_{1})\big(\frac{1}{\text{BC}_{1}} + \frac{1}{\text{DC}_{2}}\big)$
Here, $OB = – u$
$DI = + v$
$BC_1 = + R_1$
$DC_2 = – R_2$
$\Rightarrow = \frac{\text{n}_{1}}{\text{-u}} + \frac{\text{n}_{1}}{\text{v}} = (\text{n}_{2} -\text{n}_{1})\bigg(\frac{1}{\text{R}_{1}} +\frac{1}{\text{R}_{2}}\bigg)$
$\Rightarrow = \text{n}_{1}\bigg(\frac{1}{\text{v}} +\frac{1}{u}\bigg) (\text{n}_{2} -\text{n}_{1})\bigg(\frac{1}{\text{R}_{1}} +\frac{1}{\text{R}_{2}}\bigg)$
$\Rightarrow\frac{1}{\text{f}} =\bigg(\frac{\text{n}_{2}}{\text{n}_{1}} -1 \bigg)\bigg(\frac{1}{\text{R}_{1}} -\frac{1}{\text{R}_{2}}\bigg)$

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Question 315 Marks

A point object 'O' is kept in a medium of refractive index $n_1$ in front of a convex spherical surface of radius of curvature R which separates the second medium of refractive index $n_2$ from the first one, as shown in the figure. Draw the ray diagram showing the image formation and deduce the relationship between the object distance and the image distance in terms of $n_1, n_2$ and R.

hen the image formed above acts as a virtual object for concave spherical, surface separating the medium $n_2$ from $n_1 (n_2> n_1),$ draw this ray diagram and write the similar (similar to (a) relation. Hence obtain the expression for the lens maker's formula.

Answer

For small angles
$\angle\text{N0M}\cong\tan\angle \text{N0M} = \frac{\text{MN}}{\text{OM}}$
$\angle\text{NCM}\cong\tan\angle \text{NCM} = \frac{\text{MN}}{\text{MC}}$
$\angle\text{NIM}\cong\tan\angle \text{NIM} = \frac{\text{MN}}{\text{MI}}$
In $\Delta\text{NOC} , $ $\angle \text{i} = \angle\text{i}\angle\text{NOM} + \angle\text{NCM}$...........................(i)
$\therefore\angle\text{i} = \frac{\text{MN}}{\text{OM}} + \frac{\text{MN}}{\text{MC}}$
$\angle\text{r} = \angle \text{NCM} -\angle \text{NIM}$
$ =\frac{\text{MN}}{\text{MC}} -\frac{\text{MN}}{\text{MI}}$
Using Snell's Law
$n_1 \sin\ i. = n_2 \sin\ r$
For small angles $n_1 i = n_2 r$
Substituting for i and r, we get
$\frac{\text{n}_{1}}{\text{OM}} +\frac{\text{n}_{2}}{\text{MI}} = \frac{\text{n}_{2} -\text{n}_{1}}{\text{MC}}$
Here, $\text{OM} =\text{-u}, \text{MI} = + \text{v},\text{MC} = + \text{R}$
Substituting these, we get
$\Rightarrow\frac{\text{n}_{2}}{\text{v}} - \frac{\text{n}_{1}}{\text{u}} = \frac{\text{n}_{2} -\text{n}_{1}}{\text{R}}$

(Alternatively accept this Ray diagram)

Similarly relation for the surface ADC.
$\frac{-\text{n}}{\text{DI}{1}} +\frac{\text{n}_{1}}{\text{DI}} = \frac{\text{n}_{2} - \text{n}_{1}}{\text{DC}_{2}}$................................(i)
Refraction at the first surface ABC of the lens.
$\frac{\text{n}_{1}}{\text{OB}} +\frac{\text{n}_{2}}{\text{BI}_{1}} = \frac{\text{n}_{2} -\text{n}_{2}}{\text{BC}_{1}}$..............................(ii)
Adding (i)and (ii), and taking $ BI_1\cong DI_1,$ we get
$\frac{\text{n}_{1}}{\text{OB}} +\frac{\text{n}_{1}}{\text{DI}} = (\text{n}_{2} -\text{n}_{1})\big(\frac{1}{\text{BC}_{1}} + \frac{1}{\text{DC}_{2}}\big)$
Here, $OB = – u$
$DI = + v$
$BC_1 = + R_1$
$DC_2 = – R_2$
$\Rightarrow = \frac{\text{n}_{1}}{\text{-u}} + \frac{\text{n}_{1}}{\text{v}} = (\text{n}_{2} -\text{n}_{1})\bigg(\frac{1}{\text{R}_{1}} +\frac{1}{\text{R}_{2}}\bigg)$
$\Rightarrow = \text{n}_{1}\bigg(\frac{1}{\text{v}} +\frac{1}{u}\bigg) (\text{n}_{2} -\text{n}_{1})\bigg(\frac{1}{\text{R}_{1}} +\frac{1}{\text{R}_{2}}\bigg)$
$\Rightarrow\frac{1}{\text{f}} =\bigg(\frac{\text{n}_{2}}{\text{n}_{1}} -1 \bigg)\bigg(\frac{1}{\text{R}_{1}} -\frac{1}{\text{R}_{2}}\bigg)$

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Question 325 Marks

Draw a ray diagram showing the image formation by a compound microscope. Hence obtain expression for total magnification when the image is formed at infinity.
Distinguish between myopia and hypermetropia. Show diagrammatically how these defects can be corrected.

Answer

Difference between myopia and hypermetropia

Myopia	Hypermetropia
The eye ball is elongated.	The eye ball is shortened.
Person cannot see distant objects clearly.	Person cannot see near objects clearly.

Myopic eye is corrected by interposing a concave lens between eye and object.
Hypermetropia is corrected by interposing a convex lens between eye and object.

If image A 'B' is exactly at the focus of the eyepiece, then image A "B" is formed at infinity. If the object AB is very close to the focus of the objective lens of focal length $f_o$, then magnification $M_o$ by the objective lens

$\text{M}_{e} = \frac{\text{L}}{\text{f}_{o}}$
where L is tube length (or distance between lenses $L_o$ and $L_e$)
Magnification $M_e$ by the eyepiece
$\text{M}_{e} = \frac{\text{D}}{\text{f}_{e}}$
where D = Least distance of distinct vision Total magnification $m = M_oM_e$
$ = \bigg(\frac{\text{L}}{\text{f}_{o}}\bigg)\bigg(\frac{\text{D}}{\text{f}_{e}}\bigg).$

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Question 335 Marks

State Huygen’s principle. Using this principle draw a diagram to show how a plane wave front incident at the interface of the two media gets refracted when it propagates from a rarer to a denser medium. Hence verifiy Snell’s law of refraction.
When monochromatic light travels from a rarer to a denser medium, explain the following, giving reasons:
1. Is the frequency of reflected and refracted light same as the frequency of incident light?
2. Does the decrease in speed imply a reduction in the energy carried by light wave?

Answer

Law of Reflection : Let XY be a reflecting surface at which a wavefront is being incident obliquely. Let v be the speed of the wavefront and at time t = 0, the wavefront touches the surface XY at A. After time t, the point B of wavefront reaches the point B' of the surface.

According to Huygen’s principle each point of wavefront acts as a source of secondary waves. When the point A of wavefront strikes the reflecting surface, then due to presence of reflecting surface, it cannot advance further; but the secondary wavelet originating from point A begins to spread in all directions in the first medium with speed v. As the wavefront AB advances further, its points $A_1, A_2 , A_3$ etc. strike the reflecting surface successively and send spherical secondary wavelets in the first medium.

First of all the secondary wavelet starts from point A and traverses distance AA' (=vt) in first medium in time t. In the same time t, the point B of wavefront, after travelling a distance BB', reaches point B' (of the surface), from where the secondary wavelet now starts. Now taking A as centre we draw a spherical arc of radius AA' (= vt) and draw tangent A'B' on this arc from point B'. As the incident wavefront AB advances, the secondary wavelets starting from points between A and B', one after the other and will touch A'B' simultaneously. According to Huygen’s principle wavefront A'B' represents the new position of AB, i.e., A'B' is the reflected wavefront corresponding to incident wavefront AB.
Now in right-angled triangles ABB' and AA' B'
$\angle\text{ABB}' = \angle\text{AA'} B' \text{(both are equal to } 90^{o})$
side BB' = side AA' (both are equal to $v\text{t}$)
and side AB' is common
i.e., both triangles are congruent.
$\therefore\angle\text{BAB}' = \angle\text{AB}'\text{A}'$
i.e., incident wavefront AB and reflected wavefront A'B' make equal angles with the reflecting surface XY. As the rays are always normal to the wavefront, therefore the incident and the reflected rays make equal angles with the normal drawn on the surface XY, i.e.,
angle of incidence i = angle of reflection r
This is the second law of reflection.
Since AB, A'B' and XY are all in the plane of paper, therefore the perpendiculars dropped on them will also be in the same plane. Therefore we conclude that the incident ray, reflected ray and the normal at the point of incidence, all lie in the same plane. This is the first law of reflection. Thus Huygen’s principle explains both the laws of reflection.

1. If the radiation of certain frequency interact with the atoms/molecules of the matter, they start to vibrate with the same frequency under forced oscillations. Thus, the frequency of the scattered light (Under reflection and refraction) equals to the frequency of incident radiation.
2. No, energy carried by the wave depends on the amplitude of the wave, but not on the speed of the wave.

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Question 345 Marks

Define magnifying power of a telescope. Write its expression.
A small telescope has an objective lens of focal length 150 cm and an eye piece of focal length 5 cm. If this telescope is used to view a 100m high tower 3 km away, find the height of the final image when it is formed 25 cm away from the eye piece.

Answer

Magnifying power is the ratio of the angle subtended at the eye by the image to the angle subtended at the unaided eye by the object.

Expression

$\text{m} = \beta\big/\alpha = \text{f}_{o}\big/\text{f}_{e}$

or $\text{m} = \text{f}_{o} /_{ \text{f}_{e}}\bigg(1 + \frac{\text{f}_{e}}{\text{D}}\bigg)$

Using, the lens equation for objective lens,:

$\frac{1}{\text{f}_{o}} =\frac{1}{\text{v}_{o}}-\frac{1}{\text{u}_{o}}$

$ => \frac{1}{150} = \frac{1}{\text{v}_{o}} - \frac{1}{-3\times10^{5}}$

$ => \frac{1}{\text{v}_{o}} =\frac{1}{150} -\frac{1}{-3\times10^{5}} = \frac{2000-1}{3\times10^{5}}$

$ =>\text{v}_{o} = \frac{3\times10^{5}}{1999}\text{cm}$

$\approx 150 \text{cm}$

Hence, magnification due to the objective lens

$\text{m}_{o} =\frac{\text{v}_{o}}{\text{u}_{o}} = \frac{150\times10^{-2}\text{m}}{3000\text{m}}$

$\approx\frac{10^{-2}}{20} = .05\times10^{-2}$

Using lens formula for eyepiece

$\frac{1}{\text{f}_{e}} =\frac{1}{\text{v}_{e}} - \frac{1}{\text{u}_{e}}$

$ = > \frac{1}{5} = \frac{1}{-25} -\frac{1}{\text{u}_{e}}$

$ = > \frac{1}{\text{u}_{e}} =\frac{1}{-25} - \frac{1}{5} = \frac{-1-5}{25}$

$ = >\text{u}_{e} =\frac{-25}{6}\text{cm}$

$\therefore\text{ Magnification due to eyepiece} \text{ m}_{e} =\frac{-25}{-\frac{25}{6}}= 6 $

Hence, total magnification $ = > \text{m} = \text{m}_{e}\times\text{m}_{o}$

$\text{m} = 6 \times5\times10^{-4} = 30 \times10^{-4}$

Hence, size of final image

$ = 30 \times10^{-4}\times100 \text{m}$

$ = 30 \text{ cm}.$

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Question 355 Marks

How is the working of a telescope different from that of a microscope?
The focal lengths of the objective and eyepiece of a microscope are 1.25 cm and 5 cm respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment.

Answer

Working differences:

Objective of a telescope forms the image of a very far off object at, or within, the focus of its eyepiece. The microscope does the same for a small object kept just beyond the focus of its objective.
The final image formed by a telescope is magnified relative to its size as seen by the unaided eye while the final image formed by a microscope is magnified relative to its absolute size.
The objective of a telescope has large focal length & large aperture while the corresponding for a microscope have very small values.

Telescope	Microscope
Resolving power should be higher for certain magnification. Focal length of objective should be kept larger while eyepiece focal length should be small for better magnification. Objective should be of large aperture. Distance between objective and eye piece is adjusted to focus the object at infinity.	Resolving power is not so large but the magnification should be higher. Both objective and eye piece should have less focal length for better magnification. Eye piece should be of large aperture. Distance between objective and eye piece is fixed, for focusing an object the distance of the objective is changed.

Given: $\text{f}_{o} = 1.25\text{cm}$
$\text{f}_{e} = 5 \text{cm}$
Angular magnification m= 30
Now, $m = m_e \times m_o$
In normal adjustment, angular magnification of eyepiece
$\text{m}_{e} = \text{d} / \text{f}_{e} = + \frac{25}{5} = 5 $
Hence, $m_o = 6$
But $\text{m}_{o} = \frac{\text{v}_{o}}{\text{u}_{o}} = > -6 =\frac{\text{v}_{o}}{\text{u}_{o}}$
$ = > \text{v}_{o} = - 6 \text{u}_{o}$
Applying lens equation to the objective lens:
$\frac{1}{\text{f}_{o}} = \frac{1}{\text{v}_{o}} - \frac{1}{\text{u}_{o}}$
$\frac{1}{1.25} = \frac{1}{-6\text{u}_{o}} - \frac{1}{\text{u}_{o}}$
$\frac{1}{1.25} = \frac{-1-6}{6\text{u}_{o}}$
$6\text{u}_{o} = 1.25\times(-7)$
$\text{u}_{o} = \frac{-1.25\times7}{6}\text{ cm}$
$ = - 1.46\text{cm}.$

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Question 365 Marks

Obtain lens makers formula using the expression

$\frac{\text{n}_{2}}{\text{v}} - \frac{\text{n}_{1}}{\text{u}} = \frac{(\text{n}_{2} - \text{n}_{1})}{\text{R}}$

Here the ray of light propagating from a rare medium of refractive index $(n_1)$ to a denser medium of refractive index $(n_2)$ is incident on the convex side of spherical refracting surface of radius of curvature R.

Draw a ray diagram to show the image formation by a concave mirror when the object is kept between its focus and the pole. Using this diagram, derive the magnification formula for the image formed.

Answer

For refraction at the first surface
$\frac{\text{n}_{2}}{\text{v}_{1}} - \frac{\text{n}_{1}}{\text{u}} = \frac{\text{n}_{2} - \text{n}_{1}}{\text{R}_{1}}$
For the second surface, $I_1$ acts as a virtual object (located in the denser medium) whose final real image is formed in the rarer medium at I.
So for refraction at this surface, we have
$\frac{\text{n}_{1}}{\text{v}} - \frac{\text{n}_{2}}{\text{v}_{1}} = \frac{\text{n}_{1} - \text{n}_{2}}{\text{R}_{2}}$
From above two equations, $\frac{1}{\text{v}} - \frac{1}{\text{u}} = \bigg(\frac{\text{n}_{2}}{\text{n}_{1}} - 1\bigg)\bigg (\frac{1}{\text{R}_{1}} - \frac{1}{\text{R}_{2}}\bigg)$
The point, where image of an object, located at intinity is formed, is called the focus F, of the lens and the distance f gives its focal length.
So for u = $\propto,\text{v} = + \text{ f}$
$\Rightarrow\frac{1}{\text{f}} = \bigg(\frac{\text{n}_{2}}{\text{n}_{1}} - 1 \bigg)\bigg(\frac{1}{\text{R}_{1}} - \frac{1}{\text{R}_{2}}\bigg)$

$\Delta$ABP is similar to $\Delta$A'B'P
So $\frac{\text{A}'\text{B}'}{\text{AB}} = \frac{\text{B}'\text{P}}{\text{BP}}$
Now A' B' = I, AB = O, B'P = + v and BP = - u
So magnification $\text{m} = \frac{\text{I}}{\text{O}} = - \frac{\text{v}}{\text{u}}.$

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Question 375 Marks

Draw a ray diagram to show refraction of a ray of monochromatic light passing through a glass prism.

Deduce the expression for the refractive index of glass in terms of angle of prism and angle of minimum deviation.

Explain briefly how the phenomenon of total internal reflection is used in fibre optics.

Answer

From $\Delta$ MQR, $(i - r_1) + (e - r_2)= \delta$
So$ (i + e)-(r_1 + r_2)= \delta$
From $\Delta PQN r_1+ r_2+ \angle QNR =180^\circ$
Also $A + \angle QNR =180^\circ$
Thus $A= r_1+ r_2$
So $ i + e - A= \delta$
At minimum deviation, $i = e, r_1= r_2 = r$ and $\delta = \delta_{m}$
$\Rightarrow \text{i} = \frac{\text{A} + \delta_{m}}{2}$
and $\text{r} = \frac{\text{A}}{2}$
Also $\mu = \frac{\sin\text{i}}{\sin\text{r}}$
Hence $\mu = \frac{\sin\bigg(\frac{\text{A} + \delta_{m}}{2}\bigg)}{\sin\bigg(\frac{\text{A}}{2}\bigg)}$

Each optical fibre consists of a core and cladding. Refractive index of the material of the core is higher than that of cladding. When a signal, in the form of light, is directed into the optical fibre, at an angle greater than the (relevant) critical angle, it undergoes repeated total internal reflections along the length of the fibre and comes out of it at the other end with almost negligible lossof intensity.

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Question 385 Marks

Draw the labelled ray diagram for the formation of image by a compound microscope.Derive the expression for the total magnification of a compound microscope. Explain why both the objective and the eyepiece of a compound microscope must have short focal lengths.

Answer

Expression for total magnification: The (linear) magnification due to the objective, namely h’/ h, equals
$\text{m}_{\circ} = \frac{\text{h}'}{\text{h}} = \frac{\text{L}}{\text{f}_{\circ}}$
where we have used the result
$\tan\beta = \bigg(\frac{\text{h}}{f}_{\circ}\bigg) = \bigg(\frac{\text{h}'}{\text{L}}\bigg)$
$\text{L}\cong$ Distance between the second focal point of the objective and the first focal point of the eye piece = tube length of the compound microscope. [When the final image is formed at infinity, the angular magnification due to the eyepiece is
$\text{m}_{e} = \big(\text{D}/f_{e}\big)]$
Alternate Answer
The (angular) magnification $m_e$ ,due to eyepiece, when the final image is formed at the near point, is
$\text{m}_{e} = \bigg(1 + \frac{\text{D}}{f}_{e}\bigg)$
The total magnification when the image is formed at infinity, is
$\text{m} = \text{m}_{\circ}\text{m}_{e} = \bigg(\frac{\text{L}}{f}_{\circ}\bigg)\bigg(\frac{\text{D}}{f}_{e}\bigg)$
Alternate Answer
The total magnification, when the final image is formed at the near point, is
$\text{m} = \text{m}_{\circ}\text{m}_{e} = \frac{\text{L}}{f}_{\circ}\bigg(1 + \frac{\text{D}}{f}_{\circ}\bigg)$
Reason: With short focal lengths of objective and eyepiece the total magnification of the compound microscope increases.

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Question 395 Marks

Trace the rays of light showing the formation of an image due to a point object placed on the axis of a spherical surface separating the two media of refractive indices $n_1$ and $n_2$. Establish the relation between the distances of the object, the image and the radius of curvature from the central point of the spherical surface. Hence derive the expression of the lens maker’s formula.

Answer

$\tan<\text{NOM} = \frac{\text{MN}}{\text{OM}}$
$<\text{NCM} \cong \tan<\text{NCM} =\frac{\text{MN}}{\text{MC}}$
$<\text{NCM}\cong \tan<\text{NCM} =\frac{\text{MN}}{\text{MI}}$
Now, for $\bigtriangleup\text{NOC}$, i is the exterior angle, Therefore, $\text{I} = <\text{NOM} + <\text{NCM}$
$\ell = \frac{\text{MN}}{\text{OM}} + \frac{\text{MN}}{\text{MC}}$
Similary
$\text{r} = < \text{NCM} -<\text{NIM}'$$\text{i}.\text{e}..\text{r} =\frac{\text{MN}}{\text{MC}} - \frac{\text{MN}}{\text{MI}}$
Now , by Snell's law
$\text{n}_{1}\sin\text{i} = \text{n}_{2}\sin \text{r} , $
or for small angles
$\text{n}_{1}\text{i} = \text{n}_{2}\text{r}$
Substituting for i and r , we then get
$\frac{\text{n}_{1}}{\text{OM}} + \frac{\text{n}_{2}}{\text{MI}} = \frac{\text{n}_{2} - \text{n}_{1}}{\text{MC}}$
Applying the Cartesian sign convention,
$\text{MO} = - u, \text{MI} = + \text{v} , \text{MC} = + \text{R} , $ we get
$\frac{\text{n}_{2}}{v} - \frac{\text{n}_{1}}{u} = \frac{\text{n}_{2} - \text{n}_{1}}{\text{R}}$
Lens Maker’s Formula:

Applying the above result to the first interface ABC, we get,
$\frac{\text{n}_{1}}{\text{OB}} + \frac{\text{n}_{2}}{\text{BI}_{1}} = \frac{\text{n}_{2} - \text{n}_{2}}{\text{BC}_{1}}$
A similar procedure applied to the second interfaceADC gives,
$ - \frac{\text{n}_{2}}{\text{DI}_{1}} + \frac{\text{n}_{1}}{\text{DI}} = \frac{\text{n}_{2} - \text{n}_{2}}{\text{DC}_{2}}$
For a thin lens, $BI_1 = DI_1$. Adding the above two equations,we then get,
$\frac{\text{n}{1}}{\text{OB}} + \frac{\text{n}_{1}}{\text{DI}} = \big(\text{n}_{2} - \text{n}_{1}\big)\bigg(\frac{1}{\text{BC}_{1}} + \frac{1}{\text{DC}_{2}}\bigg)$
Suppose the object is at infinity. Then
$\text{OB} = \infty$ and $\text{DI} = f$ and the above equation becomes
$\frac{\text{n}_{1}}{f} = \big(\text{n}_{2} - \text{n}_{1}\big)\bigg(\frac{1}{\text{BC}_{1}} + \frac{1}{\text{DC}_{2}}\bigg)$
By the sign convention,
$\text{BC}_{1} = + \text{R}_{1}.$
$\text{DC}_{1} = - \text{R}_{2}$ $\frac{1}{f} = \big(\text{n}_{21} - 1\big)\bigg(\frac{1}{\text{R}_{1}} - \frac{1}{\text{R}_{2}}\bigg)$ $\big(\because\text{n}_{21} = \frac{\text{n}_{2}}{\text{n}_{1}}\big)$

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Question 405 Marks

For a ray of light travelling from a denser medium of refractive index $n_1$ to a rarer medium of refractive index $n_2$ prove that $\frac{\text{n}_{2}}{\text{n}_{1}} = \sin \text{i}_{c} ,$ where $i_c$ is the critical angle of incidence for the media.
Explain with the help of a diagram, how the above principle is used for transmission of video signals using optical fibres.

Answer

(i)

$\frac{\text{n}_{2}}{\text{n}_{1}} = \frac{\sin\text{i}}{\sin\text{r}}$

$\text{For } \angle\text{i} =\angle\text{i}_{c} \text{ (critical angle)}$
$\angle\text{r} = 90^{o}$
$\therefore$ $\frac{\sin \text{i}_{c}}{\sin90^{\circ}} = \frac{\text{n}_{2}}{\text{n}_{1}}$
or $\sin\text{i}_{c} = \frac{\text{n}_{2}}{\text{n}_{1}}$

(i)

(ii) When a video signal is directed at one end of the fiber at a suitable angle, it undergoes repeated total internal reflections along the length of the fiber and comes out of it.

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Question 415 Marks

Draw a ray diagram to show image formation when the concave mirror produces a real, inverted and magnified image of the object.
Obtain the mirror formula and write the expression for the linear magnification.
Explain two advantages of a reflecting telescope over a refracting telescope.

Answer

Object between F and C. The image is Beyond C.

No chromatic aberration.
Spherical aberration reduces significantly.
Low set up cost, lighter, forms brighter image.

From the concavemirror.
Using cartesian sign convention, we find

Object distance,	BP = -u
Image distance,	B'P = -v
Focal length,	FP = -f
Radius of curvature,	CP = -R = -2f

Now, $\triangle\text{A}'\text{B}'\text{C}\sim\triangle\text{ABC}$
$\therefore\ \frac{\text{A}'\text{B}'}{\text{AB}}=\frac{\text{CB}'}{\text{BC}}=\frac{\text{CP}-\text{B}'\text{P}}{\text{BP}-\text{CP}}=\frac{-\text{R}+\text{v}}{-\text{u}+\text{R}}\ ....(1)$
As $\angle\text{A}'\text{PB}'=\angle\text{APB},$ therefore,
$\triangle\text{A}'\text{B}'\text{P}\sim\triangle\text{ABP}.$
Consequently,
$\frac{\text{A}'\text{B}'}{\text{AB}}=\frac{\text{B}'\text{P}}{\text{BP}}=\frac{-\text{v}}{-\text{u}}=\frac{\text{v}}{\text{u}}\ ....(2)$
From equations (1) and (2), we get
$\frac{-\text{R}+\text{v}}{-\text{u}+\text{R}}=\frac{\text{v}}{\text{u}}$
or, $-\text{uR}+\text{uv}=\text{-uv}+\text{vR}$
or, $\text{vR}+\text{uR}=2\text{uv}$
Dividing both sides by uvR, we get
$\frac{1}{\text{u}}+\frac{1}{\text{v}}=\frac{2}{\text{R}}$
But, R = 2f
$\frac{1}{\text{u}}+\frac{1}{\text{v}}=\frac{1}{\text{f}}$
This proves themirror formula for a concavemirror, when it forms a real image.
Linear magnification is given by:
$\text{m}=\frac{\text{h}_2}{\text{h}_1}=\frac{-\text{v}}{\text{u}}$
here $h_2, h_1$ are heights of image and object respectively v, u are distances of image and object from the pole of mirror respectively.

Nature Real, Inverted and Larger than object.
Consider an object AB placed on the principal axis beyond the centre of curvature C of a concave mirror of small aperture, as shown in Fig. Aray AM from the object travels parallel to the principal axis and after reflection from the mirror it passes through focus F. Another ray AP is incident on the pole P of the mirror and is reflected along PA' in accordance with the laws of reflection so that $\angle\text{APB}=\angle\text{BPA}'.$ The two reflected rays meet at point A'. Thus A' is the real image of A. The image of any point on AB will lie on a corresponding point of A'B. Hence A'B is the real image of AB formed by reflection.

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Question 425 Marks

An object is placed in front of a concave mirror. It is observed that a virtual image is formed. Draw the ray diagram to show the image formation and hence derive the mirror equation $\frac{1}{\text{f}}=\frac{1}{\text{u}}+\frac{1}{\text{v}}.$
An object is placed 30cm in front of a plano-convex lens with its spherical surface of radius of curvature 20cm. If the refractive index of the material of the lens is 1.5, find the position and nature of the image formed.

Answer

$\triangle\text{ABC}\sim\triangle\text{A}_1\text{B}_1\text{C}$
$\Rightarrow\frac{\text{A}_1\text{B}_1}{\text{AB}}=\frac{\text{A}_1\text{C}}{\text{AC}}=\frac{(+\text{v})+(-\text{R)}}{(-\text{R})-(-\text{u)}}\ ...(1)$
$\triangle\text{ABP}\sim\triangle\text{A}_1\text{B}_1\text{P}$
$\Rightarrow\frac{\text{A}_1\text{B}_1}{\text{AB}}=\frac{\text{A}_1\text{P}}{\text{AP}}=\frac{+\text{v}}{-\text{u}}\ ...(2)$
$(1)=(2)$
$\Rightarrow\frac{\text{v}-\text{R}}{-\text{R}+\text{u}}=\frac{\text{v}}{-\text{u}}$
$\Rightarrow-\text{uv}+\text{uR}=-\text{vR}+\text{uv}$
$\Rightarrow\text{uR}+\text{vR}=2\text{uv}$
$\div\text{ by }\text{uvR}$
$\Rightarrow\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{2}{\text{R}}$
$\because\text{R}=2\text{F}$
$\therefore\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\frac{1}{\text{f}}=(1.5-1)\Big(\frac{1}{20}-\frac{1}{\infty}\Big)$

$=\frac{0.5}{20}=\frac{5}{200}=\frac{1}{40}$
$\therefore\text{f}=40\text{cm}$
Now, $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{v}}$
$\Rightarrow\text{v}=\frac{\text{fu}}{\text{f}+\text{u}}=\frac{40\times-30}{40-30}$
$\Rightarrow\text{v}=\frac{-40\times30}{10}=-120\text{cm}$
Image is virtual, erect and enlarged in front of lens 120cm away.

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Question 435 Marks

Draw the ray diagram of an astronomical telescope when the final image is formed at infinity. Write the expression for the resolving power of the telescope.
An astronomical telescope has an objective lens of focal length 20m and eyepiece of focal length 1cm.

Find the angular magnification of the telescope.
If this telescope is used to view the Moon, find the diameter of the image formed by the objective lens. Given the diameter of the Moon is $3.5 \times 10^6m$ and radius of lunar orbit is $3.8 \times 10^8m.$

Answer

Resolving power $=\frac{\text{D}}{1.22\lambda}$

$\text{m}=-\frac{\text{f}_0}{\text{f}_\text{e}}=-\frac{20}{10-2}=-2000$

$\tan\alpha=\frac{\text{d}_0}{\text{u}}=\frac{\text{di}}{\text{f}_0}$
$\Rightarrow\text{d}_1=\frac{3.5\times10^6}{3.8\times10^8}\times20=0.18\text{m}$

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Question 445 Marks

A paperweight in the form of a hemisphere of radius 3.0cm is used to hold down a printed page. An observer looks at the page vertically through the paperweight. At what height above the page will the printed letters near the centre appear to the observer?

Answer

In the first refraction at A.
$\mu_2=\frac{3}{2}, \ \mu_1=1, \ \text{u}=0, \ \text{R}=\infty$
So, $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$
$\Rightarrow \frac{\frac{3}{2}}{\text{v}}-\frac{1}{0}=\frac{\mu_2-\mu_1}{\infty}$
$\Rightarrow\text{v}=0$ since $\big(\text{R}\Rightarrow\infty \ \text{and} \ \text{u}=0\big)$
$\therefore$ The image will be formed at the point, Now for the second refraction at B,
$\frac{\mu_2}{\text{v}}-\frac{\mu}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$
$\text{u}=-3\text{cm}, \ \text{R}=-3\text{cm}, \ \mu_1=\frac{3}{2}, \ \mu_2=1$
So, $\frac{1}{\text{v}}+\frac{3}{2\times3}=\frac{1-1.5}{-3}=\frac{1}{6}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{6}-\frac{1}{2}=-\frac{1}{3}$
$\Rightarrow \text{v}=-3\text{cm},$
$\therefore$ There will be no shift in the final image.

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Question 455 Marks

A convex lens of focal length 20cm and a concave lens of focal length 10cm are placed 10cm apart with their principal axes coinciding. A beam of light travelling parallel to the principal axis and having a beam diameter 5.0mm, is incident on the combination. Show that the emergent beam is parallel to the incident one. Find the beam diameter of the emergent beam.

Answer

Let, the parallel beam is first incident on convex lens.
d = diameter of the beam = 5mm
Now, the image due to the convex lens should be formed on its focus (point B)
So, for the concave lens,
u = +10cm (since, the virtual object is on the right of concave lens)
f = -10cm
So, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{-10}+\frac{1}{10}=0\Rightarrow\text{v}=\infty$
So, the emergent beam becomes parallel after refraction in concave lens.
As shown from the triangles XYB and PQB,
$\frac{\text{PQ}}{\text{XY}}=\frac{\text{RB}}{\text{ZB}}=\frac{10}{20}=\frac{1}{2}$
So, $\text{PQ}=\frac{1}{2}\times5=2.5\text{mm}$
So, the beam diameter becomes 2.5mm.
Similarly, it can be proved that if the light is incident of the concave side, the beam diameter will be 1cm.

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Question 465 Marks

For a glass prism $(\mu=\sqrt{3})$ the angle of minimum deviation is equal to the angle of the prism. Find the angle of the prism.

Answer

We know that, $\mu=\frac{\sin\Big[\frac{(\text{A}+\text{D}_\text{m})}{2}\Big]}{\sin\Big(\frac{\text{A}}{2}\Big)}$
We are given that, the angle of minimum deviation is equal to the angle of the ptism i.e., $D_m = A$
Substituring $\mu=\sqrt{3}\text{ and D}_\text{m}=\text{A in }\mu=\frac{\sin\Big[\frac{(\text{A}+\text{D}_\text{m})}{2}\Big]}{\sin\Big(\frac{\text{A}}{2}\Big)}$, we get
$\sqrt{3}=\frac{\sin\text{A}}{\sin\frac{\text{A}}{2}}$
$\Rightarrow\ \sqrt{3}=\frac{2\sin\frac{\text{A}}{2}\cos\frac{\text{A}}{2}}{\sin\frac{\text{A}}{2}}\ \ \Big[\because\ \sin \text{A}=2\sin\frac{\text{A}}{2}\cos\frac{\text{A}}{2}\Big]$
$\Rightarrow\ \sqrt{3}=2\cos\frac{\text{A}}{2}$
$\Rightarrow\ \frac{\sqrt{3}}{2}=\cos\frac{\text{A}}{2}$
$\Rightarrow\ \frac{\text{A}}{2}=30^\circ$
$\Rightarrow\ \text{A}=60^\circ$

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Question 475 Marks

A biconvex thick lens is constructed with glass $(\mu=1.50)$ Each of the surfaces has a radius of 10cm and the thickness at the middle is 5cm. Locate the image of an object placed far away from the lens.

Answer

$\text{R}_1=\text{R}_2=10\text{cm}, \ \text{t}=5\text{cm}, \ \text{u}=-\infty$
For the first refraction, (at A)
$\frac{\mu_{\text{g}}}{\text{v}}-\frac{\mu_{\text{a}}}{\text{u}}=\frac{\mu_{\text{g}}-\mu_{\text{a}}}{\text{R}_1}$
$\Rightarrow \frac{1.5}{\text{v}}-\Big(-\frac{1}{\infty}\Big)=\frac{1.5-1}{10}$
$\Rightarrow \frac{1.5}{\text{v}}=\frac{0.5}{10}$
$\Rightarrow\text{v}=30\text{cm}$
Again, for $2^{nd}$ surface, u = (30 - 5) = 25cm (virtual object)
$R_2 = -10cm$
So, $\frac{1}{\text{v}}-\frac{15}{25}=\frac{-0.5}{-10}\Rightarrow\text{v}=9.1\text{cm.}$
So, the image is formed 9.1cm further from the $2^{nd}$ surface of the lens.

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Question 485 Marks

The refractive index of a material $M_1$ changes by 0.014 and that of another material $M_2$ changes by 0.024 as the colour of the light is changed from red to violet. Two thin prisms one made of $M_1(A = 5.3^\circ)$ and other made of $M_2(A = 3.7^\circ)$ are combined with their refracting angles oppositely directed.

Find the angular dispersion produced by the combination.
The prisms are now combined with their refracting angles similarly directed. Find the angular dispersion produced by the combination.

Answer

Given that, $\mu'_\text{v}-\mu'_\text{r}=0.014$ and $\mu_\text{v}-\mu_\text{r}=0.024$ $\text{A}' = 5.3^\circ\ \text{and A} = 3.7^\circ$

When the prisms are oppositely directed,

angular dispersion $=(\mu_\text{v}-\mu_\text{r})\text{A}-(\mu'_\text{v}-\mu'_\text{r})\text{A}'$
$=0.024\times3.7^\circ-0.014\times5.3^\circ=0.0146^\circ$

When they are similarly directed,

angular dispersion $=(\mu_\text{v}-\mu_\text{r})\text{A}+(\mu'_\text{v}-\mu'_\text{r})\text{A}'$
$=0.024\times3.7^\circ+0.014\times5.3^\circ=0.163^\circ$

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Question 495 Marks

A small object is placed at the centre of the bottom of a cylindrical vessel of radius 3cm and height 4cm filled completely with water. Consider the ray leaving the vessel through a corner. Suppose this ray and the ray along the axis of the vessel are used to trace the image. Find the apparent depth of the image and the ratio of real depth to the apparent depth under the assumptions taken. Refractive index of water = 1.33

Answer

According to the figure, $\frac{\text{x}}{3}=\cot\text{r} \ ...(1)$
Again, $\frac{\sin\text{i}}{\sin\text{r}}=\frac{1}{1.33}=\frac{3}{4}$
$\Rightarrow\sin\text{r}=\frac{4}{3}\sin\text{i}=\frac{4}{3}\times\frac{3}{5}=\frac{4}{5} \ \Big(\text{because}\sin\text{i}=\frac{\text{BC}}{\text{AC}}=\frac{3}{5}\Big)$
$\Rightarrow\cot\text{r}=\frac{3}{4} \ ...(2)$
From (1) and (2)
$\Rightarrow\frac{\text{x}}{3}=\frac{3}{4}$
$\Rightarrow\text{x}=\frac{9}{4}=2.25\text{cm}.$
$\therefore \ $ Ratio of real and apparent depth = 4 : (2.25) = 1.78

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Question 505 Marks

A concave mirror forms an image of 20cm high object on a screen placed 5.0m away from the mirror. The height of the image is 50cm. Find the focal length of the mirror and the distance between the mirror and the object.

Answer

Given that,
$H_1 = 20cm, v = -5m = -500cm, h_2 = 50cm$
Since, $\frac{-\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
or $\frac{500}{\text{u}}=-\frac{50}{20}$ (because the image in inverted)
or $\text{u}=-\frac{500\times2}{5}=-200\text{cm}=-2\text{m}$
$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$ or $\frac{1}{-5}+\frac{1}{-2}=\frac{1}{\text{f}}$
or $\text{f}=\frac{-10}{7}=-1.44\text{m}$
So, the focal length is 1.44m.

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5 Marks Questions - Physics STD 12 Science Questions - Vidyadip