5 Marks Questions - Page 2 - Physics STD 12 Science Questions - Vidyadip

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5 Marks Questions

Question 515 Marks

A gravitational lens may be assumed to have a varying width of the form
$\text{w}(\text{b})=\text{k}_1\text{ In }\Big(\frac{\text{k}_2}{\text{b}}\Big)\ \ \text{b}_\text{min}<\text{b}<\text{b}_\text{max}$
$=\text{k}_1\text{ In }\Big(\frac{\text{k}_2}{\text{b}_\text{min}}\Big)\ \ \text{b}<\text{b}_\text{min}$
Show that an observer will see an image of a point object as a ring about the center of the lens with an angular radius
$\beta=\sqrt{\frac{(\text{n}-1)\text{k}_1\frac{\text{u}}{\text{v}}}{\text{u}+\text{v}}}$

Answer

It this case, differentiating expression of time taken t w.r.t.b.
$\text{t}=\frac{1}{\text{c}}\bigg(\text{u}+\text{v}+\frac{1}{2}\frac{\text{b}^2}{\text{D}}+(\text{n}-1)\text{k}_1\text{ In }\Big(\frac{\text{k}_2}{\text{b}}\Big)\bigg)$
$\frac{\text{dt}}{\text{db}}=0=\frac{\text{b}}{\text{D}}-(\text{n}-1)\frac{\text{k}_1}{\text{b}}$
$\Rightarrow\ \text{b}^2=(\text{n}-1)\text{k}_1\text{D}$
$\therefore\ \text{b}=\sqrt{(\text{n}-1)\text{k}_1\text{D}}$
Thus, all rays passing at a height b shall contribute to the image. They ray paths make an angle.
$\beta=\frac{\text{b}}{\text{v}}=\frac{\sqrt{(\text{n}-1)\text{k}_1\text{D}}}{\text{v}}=\sqrt{\frac{(\text{n}-1)\text{k}_1\text{uv}}{\text{v}^2(\text{u}+\text{v})}}=\sqrt{\frac{(\text{n}-1)\text{k}_1\text{u}}{(\text{u}+\text{v})\text{v}}}$. This is the required expression.

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Question 525 Marks

A point source S is placed midway between two converging mirrors having equal focal length f. Find the values of d for which only one image is formed.

Answer

Both the mirrors have equal focal length f.
They will produce one image under two conditions.
Case I: When the source is at distance ‘2f’ from each mirror i.e. the source is at centre of curvature of the mirrors, the image will be produced at the same point S. So, d = 2f + 2f = 4f.
Case II: When the source S is at distance ‘f’ from each mirror, the rays from the source after reflecting from one mirror will become parallel and so these parallel rays after the reflection from the other mirror the object itself. So, only sine image is formed.
Here, d = f + f = 2f

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Question 535 Marks

A converging lens of focal length 15cm and a converging mirror of focal length 10cm are placed 50cm apart with common principal axis. A point source is placed in between the lens and the mirror at a distance of 40cm from the lens. Find the locations of the two images formed.

Answer

The object is placed in the focus of the converging mirror.
There will be two images.

One due to direct transmission of light through lens.
One due to reflection and then transmission of the rays through lens.

Case I: (S') For the image by direct transmission,
u = -40cm, f = 15cm
$\Rightarrow\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{15}+\frac{1}{-40}$
$\Rightarrow\text{v}=24\text{cm}$ (left of lens)
Case II: (S'') Since, the object is placed on the focus of mirror, after reflection the rays become parallel for the lens.
So, $\text{u}=\infty$
$\Rightarrow\text{f}=15\text{cm}$
$\Rightarrow\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow \text{v}=15\text{cm}$ (left of lens).

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Question 545 Marks

A pin of length 2.0cm lies along the principal axis of a converging lens, the centre being at a distance of 11cm from the lens. The focal length of the lens is 6cm. Find the size of the image.

Answer

Now we have to calculate the image of A and B.
Let the images be A',B'.
So, length of A'B' = size of image.
For A, u = -10cm, f = 6cm
Since, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}-\frac{1}{-10}=\frac{1}{6}$
⇒ v = 15cm = OA'
For B, u = -12cm, f = 6cm
Again, $\Rightarrow\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{6}-\frac{1}{12}$
⇒ v = 12cm = OB'
$\therefore$ A'B' = OA' - OB' = 15 - 12 = 3cm.
So, size of image = 3cm.

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Question 555 Marks

A diverging lens of focal length 20cm and a converging lens of focal length 30cm are placed 15cm apart with their principal axes coinciding. Where should an object be placed on the principal axis so that its image is formed at infinity?

Answer

Given that, $f_1 = $ focal length of converging lens $= 30cm$
$f_2 =$ focal length of diverging lens $= -20cm$
and d = distance between them $= 15cm$
Let, F = equivalent focal length
So, $\therefore \ \frac{1}{\text{F}}=\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}-\frac{\text{d}}{\text{f}_1\text{f}_2}$
$\Rightarrow\frac{1}{30}+\Big(-\frac{1}{20}\Big)-\Big(\frac{15}{30(-200)}\Big)=\frac{1}{120}$
$\Rightarrow\text{F}=120\text{cm}$
⇒ The equivalent lens is a converging one.
Distance from diverging lens so that emergent beam is parallel (image at infinity),
$\text{d}_1=\frac{\text{dF}}{\text{f}_1}=\frac{15\times120}{30}=60\text{cm}$
It should be placed 60cm left to diverging lens.
⇒ Object should be placed (120 - 60) = 60cm from diverging lens.
Similarly, $\text{d}_2=\frac{\text{dF}}{\text{f}_2}=\frac{15\times120}{20}=90\text{cm}$
So, it should be placed 90cm right to converging lens.
⇒ Object should be placed (120 + 90) = 210cm right to converging lens.

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Question 565 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 575 Marks

Will the focal length of a lens for red light be more, same or less than that for blue light?

Answer

Key concept: The refractive index depends on colour of light of wavelength of light.
Cauchy's equation: $\mu=\text{A}+\frac{\text{B}}{\lambda_2}+\frac{\text{C}}{\lambda^4}+\ ....$
As $\lambda_\text{red}>\lambda_\text{blue}\text{ hence }\mu_\text{red}<\mu_\text{blue}$
Hence parallel beams of light incidnt on a lens will be bent more towards the axis for blue light compared to red.
By lens maker's formula,
$\frac{1}{\text{f}}=(\mu-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$
The refractive index for red light is less than that for blue light, $\mu_\text{red}<\mu_\text{blue}$
Hence $\frac{1}{\text{f}_\text{red}}<\frac{1}{\text{f}_\text{blue}}\Rightarrow\ \text{f}_\text{red}>\text{f}_\text{blue}$
Thus, the focal length for red light will ne greater than that for blue light.

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Question 585 Marks

A particle executes a simple harmonic motion of amplitude 1.0cm along the principal axis of a convex lens of focal length 12cm. The mean position of oscillation is at 20cm from the lens. Find the amplitude of oscillation of the image of the particle.

Answer

When the object is at 19cm from the lens, let the image will be at,$ v_1.$
$\Rightarrow\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}_1}-\frac{1}{-19}=\frac{1}{12}$
$\Rightarrow\text{v}_1=32.57\text{cm}$
Again, when the object is at 21cm from the lens, let the image will be at, $v_2.$
$\Rightarrow\frac{1}{\text{v}_2}-\frac{1}{\text{u}_2}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}_2}-\frac{1}{21}=\frac{1}{12}$
$\Rightarrow\text{v}_2=28\text{cm}$
$\therefore$ Amplitude of vibration of the image is A $=\frac{\text{A}'\text{B}'}{2}=\frac{\text{v}_1-\text{v}_2}{2}$
$\Rightarrow\text{A}=\frac{32.57}{2}=2.285\text{cm}$

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Question 595 Marks

A converging mirror $\mathrm{M}_1$, a point source S and a diverging mirror $\mathrm{M}_2$ are arranged as shown in figure. The source is placed at a distance of 30 cm from $\mathrm{M}_1$. The focal length of each of the mirrors is 20 cm . Consider only the images formed by a maximum of two reflections. It is found that one image is formed on the source itself.

Find the distance between the two mirrors.
Find the location of the image formed by the single reflection from $\mathrm{M}_2$.

Answer

As shown in figure, for $1^{st}$ reflection in $M_1, u = -30cm, f = -20cm$

$\Rightarrow\frac{1}{\text{v}}+\frac{1}{-30}=-\frac{1}{20}\Rightarrow\text{v}=-60\text{cm}.$
So, for 2^{nd} reflection in $M_2$
$u = 60 - (30 + x) = 30 - x$
$v = -x; f = 20cm$
$\Rightarrow\frac{1}{30-\text{x}}-\frac{1}{\text{x}}=\frac{1}{20}$
$\Rightarrow \frac{\text{x}-30+\text{x}}{\text{x}(30-\text{x})}=\frac{1}{20}$
$\Rightarrow40\text{x}-600=30\text{x}-\text{x}^2$
$\Rightarrow\text{x}^2+10\text{x}-600=0$
$\Rightarrow\text{x}=\frac{10\pm50}{2}=\frac{40}{2}=20\text{cm}$ or $-30\text{cm}$
$\therefore$ Total distance between the two lines is $20 + 30 = 50cm.$

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Question 605 Marks

A double convex lens has focal length 25cm. The radius of curvature of one of the surfaces is double of the other. Find the radii, if the refractive index of the material of the lens is 1.5.

Answer

For the double convex lens
$f = 25cm, R_1= R$ and $R_2 = -2R$ (sign convention)
$\frac{1}{\text{f}}=(\mu-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$
$\Rightarrow\frac{1}{25}=(15-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{-2\text{R}}\Big)=0.5\Big(\frac{3\text{R}}{2}\Big)$
$\Rightarrow\frac{1}{25}=\frac{3}{4}\frac{1}{\text{R}}\Rightarrow\text{R}=18.75\text{cm}$
$R_1= 18.75cm, R_2 = 2R = 37.5cm.$

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Question 615 Marks

A point object is placed at a distance of 15cm from a convex lens. The image is formed on the other side at a distance of 30cm from the lens. When a concave lens is placed in contact with the convex lens, the image shifts away further by 30cm. Calculate the focal lengths of the two lenses.

Answer

Let u = object distance from convex lens = -15cm
$v_1 =$ image distance from convex lens when alone = 30cm
$f_1 =$ focal length of convex lens.
Now, $\therefore \ \frac{1}{\text{v}_1}+\frac{1}{\text{u}}=\frac{1}{\text{f}_1}$
or, $\frac{1}{\text{f}_1}=\frac{1}{30}-\frac{1}{-15}=\frac{1}{30}+\frac{1}{15}$
or, $\text{f}_1=10\text{cm}$
Again, Let v = image (final) distance from concave lens = +(30 + 30) = 60cm
$v_1 =$ object distance from concave lens = +30m
$f_2 =$ focal length of concave lens
Now, $\therefore \ \frac{1}{\text{v}}-\frac{1}{\text{v}_1}=\frac{1}{\text{f}_1}$
Or, $=\frac{1}{\text{f}_1}=\frac{1}{60}-\frac{1}{30}\Rightarrow\text{f}_2=-60\text{cm}.$
So, the focal length of convex lens is 10cm and that of concave lens is 60cm.

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Question 625 Marks

A lady uses +1.5D glasses to have normal vision from 25cm onwards. She uses a 20D lens as a simple microscope to see an object. Find the maximum magnifying power if she uses then microscope

Together with her glass
Without the glass. Do the answers suggest that an object can be more clearly seen through a microscope without using the correcting glasses?

Answer

The lady uses +1.5D glasses to have normal vision at 25cm.So, with the glasses, her least distance of clear vision = D = 25cm
Focal length of the glasses $=\frac{1}{1.5}\text{m}=\frac{100}{1.5}\text{cm}$
So, without the glasses her least distance of distinct vision should be more
If, $\text{u}=-25\text{cm},$
$\text{f}=\frac{100}{1.5}\text{cm}$
Now, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}=\frac{1.5}{100}-\frac{1}{25}=\frac{1.5{-4}}{100}=\frac{-2.5}{100}$
$\Rightarrow \text{v}=-40\text{cm}$ = near point without glasses
Focal length of magnifying glass $=\frac{1}{20}\text{m}=0.05\text{m}=5\text{cm}=\text{f}$

The maximum magnifying power with glasses

$\text{m} =1+\frac{\text{D}}{\text{f}}=1+\frac{25}{5}=6$

Without the glasses, D = 40cm.

So, $\text{m}=1+\frac{\text{D}}{\text{f}}=1+\frac{40}{5}=9$

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Question 635 Marks

A cylindrical vessel of diameter 12cm contains $800\pi\text{cm}^3$ of water. A cylindrical glass piece of diameter 8.0 cm and height 8.0cm is placed in the vessel. If the bottom of the vessel under the glass piece is seen by the paraxial rays, locate its image. The index of refraction of glass is 1.50 and that of water is 1.33.

Answer

Given $r = 6 cm, r_1 = 4cm, h_1 = 8cm$
Let, h = final height of water column.
The volume of the cylindrical water column after the glass piece is put will be,
$\pi\text{r}^2\text{h}=800\pi+\pi\text{r}_1{^2\text{h}_1}$
or $\text{r}^2\text{h}=800+\text{r}_1{^2\text{h}_1}$
or $6^2\text{h}=800+4^2\times8$
$\text{h}=\frac{800+128}{36}=\frac{928}{36}$
$=25.7\text{cm}$
There are two shifts due to glass block as well as water.
So, $\Delta\text{t}_1=\Big(1-\frac{1}{\mu_0}\Big)\text{t}_0=\bigg(1-\frac{1}{\frac{3}{2}}\bigg)8=2.26\text{cm}$
And, $\Delta\text{t}_2=\Big(1-\frac{1}{\mu_{\text{w}}}\Big)\text{t}_{\text{w}}=\bigg(1-\frac{1}{\frac{4}{3}}\bigg)(25.7-8)=4.44\text{cm}$
Total shift = (2.66 + 4.44)cm = 7.1cm above the bottom.

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Question 645 Marks

The radii of curvature of a lens are +20cm and +30cm. The material of the lens has a refracting index 1.6. Find the focal length of the lens.

If it is placed in air.
If it is placed in water $(\mu=1.33).$

Answer

$\text{R}_1=+20\text{cm}; \ \text{R}_2=+30\text{cm}; \ \mu=1.6$

If placed in air:

$\frac{1}{\text{f}}=(\mu_{\text{g}}-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)=\Big(\frac{1.6}{1}-1\Big)\Big(\frac{1}{20}-\frac{1}{30}\Big)$

$\Rightarrow\text{f}=\frac{60}{6}=100\text{cm}$

If placed in water:

$\frac{1}{\text{f}}=(\mu_{\text{w}}-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$

$\Rightarrow \Big(\frac{1.6}{1.33}-1\Big)\Big(\frac{1}{20}-\frac{1}{30}\Big)$

$\Rightarrow \Big(\frac{1.60}{1.33}-1\Big)\Big[\frac{1}{6}\Big]$

$=\frac{28}{133\times60}\simeq\frac{1}{300}$

$\Rightarrow\text{f}=300\text{cm}$

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Question 655 Marks

A ball is kept at a height h above the surface of a heavy transparent sphere made of a material of refractive index $\mu.$ The radius of the sphere is R. At t = 0, the ball is dropped to fall normally on the sphere. Find the speed of the image formed as a function of time for $\text{t}<\sqrt{\frac{2\text{h}}{\text{g}}}.$ Consider only the image by a single refraction.

Answer

Let us assume that it has taken time ‘t’ from A to B.
$\therefore \ \text{AB}=\frac{1}{2}\text{gt}^2$
$\therefore \ \text{BC}= \text{h}-\frac{1}{2}\text{gt}^2$
This is the distance of the object from the lens at any time ‘t’.
Here, $\text{u}=-\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)$
$\mu_2=\mu$ (given) and $\mu_1=\text{i}$ (air)
So, $\Rightarrow\frac{\mu}{\text{v}}=\frac{1}{-\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)}=\frac{\mu-1}{\text{R}}$
$\Rightarrow\frac{\mu}{\text{v}}=\frac{\mu-1}{\text{R}}=\frac{1}{\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)}=\frac{(\mu-1)\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)-\text{R}}{\text{R}\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)}$
So, v = image distance at any time ‘t $=\frac{\mu\text{R}\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)}{(\mu-1)\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)-\text{R}}$
So, velocity of the image $=\text{V}=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\begin{bmatrix}\frac{\mu\text{R}\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)}{(\mu-1)\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)-\text{R}} \end{bmatrix}=\frac{\mu\text{R}^2\text{gt}}{(\mu-1)\Big(\text{h}-\frac{1}{2}\text{gt}^2\Big)-\text{R}}$ (can be found out).

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Question 665 Marks

A U-shaped wire is placed before a concave mirror having radius of curvature 20cm. Find the total length of the image.

Answer

$\text{R}=20\text{cm}, \ \text{f}=\frac{\text{R}}{2}=-10\text{cm}$ For part AB, PB = 30 + 10 = 40cm So, $\text{u}=-40\text{cm}$$\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}=-\frac{1}{10}-\Big(\frac{1}{-40}\Big)=-\frac{3}{40}$
$\Rightarrow\text{v}=-\frac{40}{3}=-13.3\text{cm}$ So, PB' = 13.3cm $\text{m}=\frac{\text{A}'\text{B}'}{\text{AB}}=-\Big(\frac{\text{v}}{\text{u}}\Big)=-\Big(\frac{-13.3}{-40}\Big)=-\frac{1}{3}$ $\Rightarrow\text{A}'\text{B}'=-\frac{10}{3}=-3.33\text{cm}$ For part CD, PC = 30, So, u = -30cm $\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}=-\frac{1}{10}+\frac{1}{-30}=-\frac{1}{15}$ $\Rightarrow\text{v}=-15\text{cm}=\text{PC}'$ So, $\text{m}=\frac{\text{C}'\text{D}'}{\text{CD}}=-\frac{\text{v}}{\text{u}}=-\Big(\frac{-15}{-30}\Big)=-\frac{1}{2}$ $\Rightarrow\text{C}'\text{D}'=5\text{cm}$ B'C' = PC' - PB' = 15 - 13.3 = 17cm So, total length A'B' + B'C' + C'D' = 3.3 + 1.7 + 5 = 10cm.

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Question 675 Marks

A short object of length L is placed along the principal axis of a concave mirror away from focus. The object distance is u. If the mirror has a focal length f, what will be the length of the image? You may take $L < < |v - f|.$

Answer

Thin mirror formula: $\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
u = object distance and v = image distance.
Since, the object distance is u. Let us consider the two ends of the object be at distance $u_1$ and $u_2$ respectively,
so that du $= |u_1 - u_2| = L$. Hence size of image can be written as dv.
By differentiating Both sides $\Big(-\frac{\text{dv}}{\text{v}^2}\Big)+\Big(-\frac{\text{du}}{\text{u}^2}\Big)=0\Rightarrow\ \frac{\text{dv}}{\text{v}^2}=-\frac{\text{du}}{\text{u}^2}$
or $\text{dv}=-\Big(\frac{\text{v}^2}{\text{u}^2}\Big)\text{du}\ .....(\text{i})$
As $\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}\Rightarrow\ \text{v}=\frac{\text{fu}}{\text{u}-\text{f}}\Rightarrow\ \text{v}=\frac{\text{fu}}{\text{u}-\text{f}}$
Or $\frac{\text{v}}{\text{u}}=\frac{\text{f}}{\text{u}-\text{f}}\ .....(\text{ii})$
From (i) and (ii) $\text{dv}=-\Big(\frac{\text{f}}{\text{u}-\text{f}}\Big)^2\text{du}$
But du = L, hence the length of the image is $\frac{\text{f}^2}{(\text{u}-\text{f})^2}\text{L}$
This is the required expression for lenght of image.

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Question 685 Marks

A normal eye has retina 2cm behind the eye-lens. What is the power of the eye-lens when the eye is

Fully relaxed,
Most strained?

Answer

Since, the retina is 2cm behind the eye-lensv = 2cm

When the eye-lens is fully relaxed

$\text{u}=\infty,$
$\text{v}=2\text{cm}=0.02\text{m}$
$\Rightarrow\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{0.02}-\frac{1}{\infty}=50\text{D}$
So, in this condition power of the eye-lens is 50D

When the eye-lens is most strained,

$\text{u}=-25\text{cm}=-0.25\text{m}$
$\text{v}=+2\text{cm}=+0.02\text{m}$
$\Rightarrow\frac{1}{\text{u}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{0.02}-\frac{1}{-0.25}=50+4=54\text{D}$
In this condition power of the eye lens is 54D.

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Question 695 Marks

A pin of length 2.00cm is placed perpendicular to the principal axis of a converging lens. An inverted image of size 1.00cm is formed at a distance of 40.0cm from the pin. Find the focal length of the lens and its distance from the pin.

Answer

Given that,
(-u) + v = 40cm = distance between object and image
$h_0 = 2cm, h_i = 1cm$
Since $\frac{\text{h}_\text{i}}{\text{h}_0}=\frac{\text{v}}{-\text{u}}=$ magnification
$\Rightarrow \frac{1}{2}=\frac{\text{v}}{-\text{u}}\Rightarrow\text{u}=-2\text{v} \ ...(1)$
Now, $\Rightarrow\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}+\frac{1}{2\text{v}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{3}{2\text{v}}=\frac{1}{\text{f}}\Rightarrow\text{f}=\frac{2\text{v}}{3} \ ...(2)$
Again, $(-\text{u})+\text{v}=40$
$\Rightarrow3\text{v}=40\Rightarrow\text{v}=\frac{40}{3}\text{cm}$
$\therefore \ \text{f}=\frac{2\times40}{3\times3}=8.89\text{cm}=$ focal length
From eqn. (1) and (2)
$u = -2v = -3f = -3(8.89) = 26.7cm =$ object distance.

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Question 705 Marks

A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5cm and the tube length is 6.5cm. Find the focal length of the eyepiece.

Answer

For the give compound microscope, $f_0 = 0.5cm$, tube length = 6.5cm magnifying power = 100 (normal adjustment) Since, the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eye piece.So, $v_o + f_e = 6.5cm ...(1)$
Again, magnifying power $\text{m}=\frac{\text{v}_0}{\text{u}_0}\times\frac{\text{D}}{\text{f}_\text{e}}$ [for normal adjustment]
$\Rightarrow\text{m}=\Big[1-\frac{\text{V}_0}{\text{f}_0}\Big]\frac{\text{D}}{\text{f}_\text{e}}$
$\Rightarrow100=-\Big[1-\frac{\text{V}_0}{0.5}\Big]\times\frac{25}{\text{f}_\text{e}}$ [Taking D = 25cm]
$\Rightarrow100\text{f}_\text{e}=-(1-2\text{v}_0)\times25$
$\Rightarrow2\text{v}_0-4\text{f}_\text{e}=1 $ ...(2)
Solving equation (1) and (2) we can get,
$V_0 = 4.5cm$ and $f_e = 2cm$
So, the focal length of the eye piece is 2cm.

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Question 715 Marks

Light is incident from glass $(\mu=1.50)$ to water $(\mu=1.33).$ Find the range of the angle of deviation for which there are two angles of incidence.

Answer

$\mu_\text{g}=1.5=\frac{3}{2}; \ \mu_\text{w}=1.33=\frac{4}{3}$For two angles of incidence,

When light passes straight through normal,

⇒ Angle of incidence = 0°, angle of refraction = 0°, angle of deviation = 0

When light is incident at critical angle,

$\frac{\sin\theta_\text{C}}{\sin\text{r}}=\frac{\mu_{\text{w}}}{\mu_{\text{g}}}$ (since light passing from glass to water)
$\Rightarrow\sin\theta_\text{C}=\frac{8}{9}$
$\Rightarrow\sin^{-1}\Big(\frac{8}{9}\Big)=62.73^{\circ}$
$\therefore$ Angle of deviation $= 90^{\circ} - \theta_\text{C} = 90^{\circ}=\sin^{-1}\Big(\frac{8}{9}\Big)=\cos^{-1}\Big(\frac{8}{9}\Big)=37.27^{\circ}$
Here, if the angle of incidence is increased beyond critical angle, total internal reflection occurs and deviation decreases. So, the range of deviation is 0 to $\cos^{-1}\Big(\frac{8}{9}\Big).$

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Question 725 Marks

A particle is moving at a constant speed V from a large distance towards a concave mirror of radius R along its principal axis. Find the speed of the image formed by the mirror as a function of the distance x of the particle from the mirror.

Answer

Given that, u = distance of the object = -x
f = focal length $=-\frac{\text{R}}{2}$
and, V = velocity of object $=\frac{\text{dx}}{\text{dt}}$
From mirror equation, $\frac{1}{-\text{x}}+\frac{1}{\text{v}}=-\frac{2}{\text{R}}$
$\frac{1}{\text{v}}=-\frac{2}{\text{R}}+\frac{1}{\text{x}}=\frac{\text{R}-2\text{x}}{\text{R}\text{x}}\Rightarrow\text{v}=\frac{\text{Rx}}{\text{R}-2\text{x}}=$ Image distance
So, velocity of the image is given by,
$\text{V}_1=\frac{\text{dv}}{\text{dt}}=\frac{\Big[\frac{\text{d}}{\text{dt}}(\text{xR})(\text{R}-2\text{x})\Big]-\Big[\frac{\text{d}}{\text{dt}}(\text{R}-2\text{x})(\text{xR})\Big]}{(\text{R}-2\text{x})^2}$
$=\frac{\text{R}\Big[\frac{\text{dx}}{\text{dt}}(\text{R}-2\text{x})\Big]-\Big[-2\frac{\text{d}}{\text{dt}}\text{x}\Big]}{(\text{R}-2\text{x})^2}=\frac{\text{R}[\text{v}(\text{R}-2\text{x})+2\text{vx}0}{(\text{R}-2\text{x})^2}$
$=\frac{\text{VR}^2}{(2\text{x}-\text{R})^2}=\frac{\text{R}\big[\text{VR}-2\text{xV}\big)+2\text{xV}}{(\text{R}-2\text{x})^2}$

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Question 735 Marks

A hemispherical portion of the surface of a solid glass sphere $(\mu=1.5)$ of radius r is silvered to make the inner side reflecting. An object is placed on the axis of the hemisphere at a distance 3r from the centre of the sphere. The light from the object is refracted at the unsilvered part, then reflected from the silvered part and again refracted at the unsilvered part. Locate the final image formed.

Answer

As shown in the figure, OQ = 3r, OP = r
So, PQ = 2r
For refraction at APB
We know, $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$
$\Rightarrow\frac{1.5}{\text{v}}-\Big(\frac{1}{-2\text{r}}\Big)=\frac{0.5}{\text{r}}=\frac{1}{2\text{r}}$ [because u = -2r]
$\Rightarrow\text{v}=\infty$
For the reflection in concave mirror
$\text{u}=\infty$
So, v = focal length of mirror $=\frac{\text{r}}{2}$
For the refraction of APB of the reflected image.
Here, $\text{u}=\frac{-3\text{r}}{2}$
$\frac{1}{\text{v}}-\frac{1.5}{\frac{-3\text{r}}{2}}=\frac{-0.5}{-\text{r}}$ $\big[\text{Here}, \ \mu_1=1.5 \ \text{and} \ \text{R}=-\text{r}\big]$
$\Rightarrow\text{v}=-2\text{r}$
As, negative sign indicates images are formed inside APB. So, image should be at C. So, the final image is formed on the reflecting surface of the sphere.

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Question 745 Marks

Figure shows a plane mirror M placed at a distance of 10cm from a concave lens L. A point object is placed at a distance of 60cm from the lens. The image formed due to refraction by the lens and reflection by the mirror is 30cm behind the mirror. What is the focal length of this lens?

Answer

The rays coming from O,
diverge on passing through concave lens,
get reflected from the mirror, would form image at $I_1 $- such that
$MI_1 = MI_2 = 30cm$
$CI_2 = MI_2 - ML = 30 - 10 = 20cm$
$I_2 $ is the position of object for plane mirror. Let $ I_2$
also be the position of virtual image of object formed by concave lens alone.

For the lens u = -60cm, v = -20cm
Using $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$ $\Rightarrow\frac{1}{-20}+\frac{1}{60}=\frac{1}{\text{f}}$ $\Rightarrow\frac{3+1}{60}=\frac{1}{\text{f}}$ $\Rightarrow\text{f}=-30\text{cm}.$

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Question 755 Marks

A converging lens of focal length 15cm and a converging mirror of focal length 10cm are placed 50cm apart. If a pin of length 2.0cm is placed 30cm from the lens farther away from the mirror, where will the final image form and what will be the size of the final image?

Answer

Given that, $f_1 = 15cm, F_m = 10cm, h_o = 2cm$
The object is placed 30cm from lens $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}.$
$\Rightarrow\text{v}=\frac{\text{uf}}{\text{u}+\text{f}}$
Since, u = -30cm and f = 15cm
So, v = 30cm
So, real and inverted image (A'B') will be formed at 30cm from the lens and it will be of same size as the object. Now, this real image is at a distance 20cm from the concave mirror. Since, $f_m = 10cm$, this real image is at the centre of curvature of the mirror. So, the mirror will form an inverted image A''B'' at the same place of same size.
Again, due to refraction in the lens the final image will be formed at AB and will be of same size as that of object. (A''B'').

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Question 765 Marks

An optical instrument used for angular magnification has a 25D objective and a 20D eyepiece. The tube length is 25cm when the eye is least strained.

Whether it is a microscope or a telescope?
What is the angular magnification produced?

Answer

The optical instrument has$\text{f}_0=\frac{1}{25\text{D}}=0.04\text{m}=4\text{cm}$
$\text{f}_\text{e}=\frac{1}{20\text{D}}=0.05\text{m}=5\text{cm}$
tube length = 25cm (normal adjustment)

The instrument must be a microscope as $f_0< f_e$
Since the final image is formed at infinity, the image produced by the objective should lie on the focal plane of the eye piece.

So, image distance for objective $= v_0 = 25 – 5 = 20cm$
Now, using lens formula.
$\frac{1}{\text{v}_0}-\frac{1}{\text{u}_0}=\frac{1}{\text{f}_0}$
$\Rightarrow\frac{1}{\text{u}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{f}_0}=\frac{1}{20}-\frac{1}{4}$
$=\frac{-4}{20}=\frac{-1}{5}\Rightarrow\text{u}_0=-5\text{cm}$
So, angular magnification $\text{m}=-\frac{\text{v}_0}{\text{u}_0}\times\frac{\text{D}}{\text{f}_\text{e}}$[Taking D = 25cm]
$=-\frac{20}{-5}\times\frac{25}{5}=20$

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Question 775 Marks

A thin convex lens of focal length 25cm is cut into two pieces 0.5cm above the principal axis. The top part is placed at (0, 0) and an object placed at (-50cm, 0). Find the coordinates of the image.

Answer

Key concept: If a symmetric lens is cut parallel to principal axis in two parts. Focal length remains the same for each part. Intensity of image formed by each part will be less compared as that of complete lens. If there was no cut, then the object would have been at a height of 0.5cm from the principal axis OO'.

The top part is placed at (0, 0) and an object placed at (-50cm, 0). There is no effect on the lenght of the lens.

Applying lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$ $\frac{1}{\text{v}}=\frac{1}{\text{u}}+\frac{1}{\text{f}}=\frac{1}{-50}+\frac{1}{25}=\frac{1}{50}\Rightarrow\ \text{v}=50\text{cm}$ Magnification is $\text{m}=\frac{\text{v}}{\text{u}}=-\frac{50}{50}=-1$ Hence the image would have been formed at 50cm from the pole and 0.5cm below the principal axis. Hence, with respect to the X-axis passing through the edge of the cut lens, the coordinates of the image are (50cm, -1cm).

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Question 785 Marks

A compound microscope consists of an objective of focal length 1cm and an eyepiece of focal length 5cm. An object is placed at a distance of 0.5cm from the objective. What should be the separation between the lenses so that the microscope projects an inverted real image of the object on a screen 30cm behind the eyepiece?

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Question 795 Marks

Consider a thin lens placed between a source (S) and an observer (O) (Fig). Let the thickness of the lens vary as $\text{w}\text{(b)}=\text{w}_0-\frac{\text{b}^2}{\alpha}$, where b is the verticle distance from the pole. $w_0$ is a constant. Using Fermat’s principle i.e. the time of transit for a ray between the source and observer is an extremum, find the condition that all paraxial rays starting from the source will converge at a point O on the axis. Find the focal length.

Answer

The time taken by ray to travel from S to $P_1$ is
$\text{t}_1=\frac{\text{SP}_1}{\text{c}}=\frac{\sqrt{\text{u}^2+\text{b}^2}}{\text{c}}$
or $\text{t}_1=\frac{\text{u}}{\text{c}}\Big(1+\frac{1}{2}\frac{\text{b}^2}{\text{u}^2}\Big)$ assuming b < < 1.
The time required to travel from $P_1$ to O is
$\text{t}_2=\frac{\text{P}_1\text{O}}{\text{c}}=\frac{\sqrt{\text{v}^2+\text{b}^2}}{\text{c}}=\frac{\text{v}}{\text{c}}\Big(1+\frac{1}{2}\frac{\text{b}^2}{\text{v}^2}\Big)$
The time required to travel through the lens is
$\text{t}=\frac{(\text{n}-1)\text{w}(\text{b})}{\text{c}}$
where n is the refractive index.
Thus, the total time is
$\text{t}=\frac{1}{\text{c}}\bigg[\text{u}+\text{v}+\frac{1}{2}\text{b}^2\Big(\frac{1}{\text{u}}+\frac{1}{\text{v}}\Big)+(\text{n}-1)\text{w}(\text{b})\bigg]$
Put $\frac{1}{\text{D}}=\frac{1}{\text{u}}+\frac{1}{\text{v}}$
Then, $\text{t}=\frac{1}{\text{c}}\bigg[\text{u}+\text{v}+\frac{1}{2}\frac{\text{b}^2}{\text{D}}+(\text{n}-1)\Big(\text{w}_0+\frac{\text{b}^2}{\alpha}\Big)\bigg]$
Fermat's principle gives the time taken should be minimum.
For that first dericative should be zero.
$\frac{\text{dt}}{\text{db}}=0=\frac{\text{b}}{\text{CD}}-\frac{2(\text{n}-1)\text{b}}{\text{c}\alpha}$
Thus, a convergent lens is formed if $\alpha=2(\text{n}-1)\text{D}$. This is independent of and hence all paraxial rays from S will converge at O i.e., for rays b < < n and b < < v.
Since, $\frac{1}{\text{D}}=\frac{1}{\text{u}}+\frac{1}{\text{v}}$, the focal length is D.

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Question 805 Marks

An astronomical telescope is to be designed to have a magnifying power of 50 in normal adjustment. If the length of the tube is 102cm, find the powers of the objective and the eyepiece.

Answer

For the astronomical telescope in normal adjustment.Magnifying power = m = 50, length of the tube = L = 102cm
Let $f_0$ and fe be the focal length of objective and eye piece respectively.
$\text{m}=\frac{\text{f}_0}{\text{f}_\text{e}}=50\Rightarrow\text{f}_0=50\text{f}_\text{e}\dots(1)$
and, $\text{L}=\text{f}_0+\text{f}_\text{e}=102\text{cm}\dots(2)$
Putting the value of f0 from equation (1) in (2), we get,
$f_0 + f_e= 102 \Rightarrow 51f_e = 102 \Rightarrow f_e= 2cm = 0.02m$
So, $f_0 = 100cm = 1m$
$\therefore$ Power of the objective lens$=\frac{1}{\text{f}_0}=1\text{D}$
And Power of the eye piece lens$=\frac{1}{\text{f}_\text{e}}=\frac{1}{0.02}=50\text{D}$

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Question 815 Marks

An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the lens is reversed?

Answer

Key concept: Thin lens formula: $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$ For a given object position if focal length of the lens deos not change, the image posrition remains unchanged.

By lens maker's formula, $\frac{1}{\text{f}}=(\mu-1)\Big(\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}\Big)$ For this position $R_1$ is positive and $R_2$ is negative.
Hence focal lenght at this position $\frac{1}{\text{f}_1}=(\mu-1)\Big(\frac{1}{(+\text{R}_1)}-\frac{1}{(-\text{R}_2)}\Big)=(\mu-1)\Big(\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}\Big)$
Now the lens is reversed,
At this position, $R_2$ is positive and $R_1$ is negative. Hence focal length at this position is $\frac{1}{\text{f}_2}=(\mu-1)\Big(\frac{1}{(+\text{R}_2)}-\frac{1}{(-\text{R}_1)}\Big)=(\mu-1)\Big(\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}\Big)$ We can observe the focal length of the lens does not change in both positions, hence the image position remains unchanged.

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Question 825 Marks

Two convex lenses, each of focal length 10cm, are placed at a separation of 15cm with their principal axes coinciding,

Show that a light beam coming parallel to the principal axis diverges as it comes out of the lens system.
Find the location of the virtual image formed by the lens system of an object placed far away.
Find the focal length of the equivalent lens.

(Note that the sign of the focal length is positive although the lens system actually diverges a parallel beam incident on it).

Answer

The beam will diverge after coming out of the two convex lens system because, the image formed by the first lens lies within the focal length of the second lens.
For $1^{st}$ convex lens, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{10} \ (\text{since, u}=-\infty)$

or, $\text{v}=10\text{cm}$
for $2^{nd}$ convex lens, $\frac{1}{\text{v}'}=\frac{1}{\text{f}}+\frac{1}{\text{u}}$
or, $\frac{1}{\text{v}'}=\frac{1}{10}+\frac{1}{-(15-10)}=\frac{-1}{10}$
or, $\text{v}'=-10\text{cm}$
So, the virtual image will be at 5cm from $1^{st}$ convex lens.

If, F be the focal length of equivalent lens,

Then, $\frac{1}{\text{F}}=\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}-\frac{\text{d}}{\text{f}_1\text{f}_2}\Rightarrow\frac{1}{10}+\frac{1}{10}-\frac{15}{100}=\frac{1}{20}$
$\Rightarrow\text{F}=20\text{cm}.$

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Question 835 Marks

Find the diameter of the image of the moon formed by a spherical concave mirror of focal length 7.6m. The diameter of the moon is 3450km and the distance between the earth and the moon is $3.8 \times 10^5km.$

Answer

$u = -3.8 \times 10^5km$
diameter of moon $= 3450km ; f = -7.6m$
$\therefore \ \frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}+\Big(-\frac{1}{3.8\times10^5}\Big)=\Big(-\frac{1}{7.6}\Big)$
Since, distance of moon from earth is very large as compared to focal length it can be taken as $\infty.$
⇒ Image will be formed at focus, which is inverted.
$\Rightarrow\frac{1}{\text{v}}=-\Big(\frac{1}{7.6}\Big)\Rightarrow\text{v}=-7.6\text{m}$
$\text{m}=-\frac{\text{v}}{\text{u}}=\frac{\text{d}_{\text{image}}}{\text{d}_{\text{object}}}\Rightarrow\frac{-(-7.6)}{(-3.8\times10^8)}=\frac{\text{d}_{\text{image}}}{3450\times10^3}$
$\text{d}_{\text{image}}=\frac{3450\times7.6\times10^3}{3.8\times10^8}=0.069\text{m}=6.9\text{cm}.$

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Question 845 Marks

A right angled prism of refractive index n has a plane of refractive index $n_1$ so that $n_1 < n$, cemented to its diagonal face. The assembly is in air. A ray is incident on AB.

Calculate the angle of incidence at AB for which the ray strikes the diagonal face at the critical angle.
Assuming n = 1.352, calculate the angle of incidence at AB for which the refracted ray passes through the diagonal face undeviated.

Answer

$\sin \text{C}=\frac{\text{n}_1}{\text{n}_2}$

$\text{In }\triangle\text{AEF}$
$(90-\text{r}_1)+45+(9-\text{C})=180$
$\Rightarrow\text{r}_1=45-\text{C}$
$\frac{\sin\text{i}}{\sin\text{r}_1}=\text{n}\sin\text{i}=\text{n}\sin\text{r}_1$
$=\text{n}\sin(45-\text{C})$
$=\text{n}(\sin45\cos\text{C}-\cos45\sin\text{C})$
$=\frac{\text{n}}{\sqrt{2}}(\cos\text{C}-\sin\text{C})$
$=\frac{\text{n}}{\sqrt{2}}\Big(\sqrt{[1-\sin^2\text{C}}]\Big)-\sin\text{C}$
$=\frac{1}{\sqrt{2}}\Big(\text{n}^2-\text{n}^2_1\Big)-\text{n}-1$
$\text{i}=\sin^{-1}\Big(\frac{1}{\sqrt{2}}\sqrt{\text{n}^2-\text{n}^2_1-\text{n}_2}\Big)$

From $\triangle\text{ ABC}$

$\text{r}_1+90^\circ+45^\circ=180^\circ$
$\Rightarrow\text{r}_1=45^\circ\frac{\sin\text{r}}{\sin\text{r}_1}$
$\sin\text{i}=\text{n}\sin\text{r}_1$
$=1.352\sin45=0.956$
$\text{i}=\sin^-1(0.956)=72.58$

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Question 855 Marks

Figure shows a transparent hemisphere of radius 3.0cm made of a material of refractive index 2.0.

A narrow beam of parallel rays is incident on the hemisphere as shown in the figure. Are the rays totally reflected at the plane surface?
Find the image formed by the refraction at the first surface.
Find the image formed by the reflection or by the refraction at the plane surface.
Trace qualitatively the final rays as they come out of the hemisphere.

Answer

Given, $\mu_2=2.0$
So, critical angle $=\sin^{-1}\Big(\frac{1}{\mu_2}\Big)=\sin^{-1}\Big(\frac{1}{2}\Big)=30^{\circ}$

As angle of incidence is greater than the critical angle, the rays are totally reflected internally.
Here, $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$

$\Rightarrow \ \frac{2}{\text{v}}-\Big(-\frac{1}{\infty}\Big)=\frac{2-1}{3}$ $\big[$ For parallel rays, $\text{u}=\infty\big]$

$\Rightarrow \ \frac{2}{\text{v}}=\frac{1}{3}\Rightarrow \text{v}=6\text{cm}$

$\Rightarrow$ If the sphere is completed, image is formed diametrically opposite of A.

Image is formed at the mirror in front of A by internal reflection.

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Question 865 Marks

A circular disc of radius 'R' is placed co-axially and horizontally inside an opaque hemispherical bowl of radius 'a' (Fig). The far edge of the disc is just visible when viewed from the edge of the bowl. The bowl is filled with transparent liquid of refractive index μ and the near edge of the disc becomes just visible. How far below the top of the bowl is the disc placed?

Answer

From the given figure, we have
AM is the direction of incidence ray before liquid is filled. After filling the bowl with the liduid, BM is the direction of the incidnet ray, Refracted ray in both cases is same as that along AM.

Let the disc is separated by O at a distance d as shown in figure.
From the figure, we have
N = 90º, OM = a, CB = NB = a - R, AN = a + R
Now, $\sin\text{t}=\frac{\text{a}-\text{R}}{\sqrt{\text{d}^2+(\text{a}-\text{R})^2}}\text{ and }\sin\alpha=\cos(90^\circ-\alpha)=\frac{\text{a}+\text{R}}{\sqrt{\text{d}^2+(\text{a}-\text{R})^2}}$
According to Snell's law we, have
$\frac{1}{\mu}=\frac{\sin\text{i}}{\sin\text{r}}=\frac{\sin\text{i}}{\sin\alpha}$
Substituting $\sin\text{i}=\frac{\text{a}-\text{R}}{\sqrt{\text{d}^2+(\text{a}-\text{R})^2}}\text{ and }\sin\alpha=\frac{\text{a}+\text{R}}{\sqrt{\text{d}^2+(\text{a}-\text{R})^2}}$, we get
$\text{d}=\frac{\mu(\text{a}^2-\text{R}^2)}{\sqrt{(\text{a}+\text{R})^2-\mu(\text{a}-\text{R})^2}}$

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Question 875 Marks

A professor reads a greeting card received on his 50th birthday with +2.5D glasses keeping the card 25cm away. Ten years later, he reads his farewell letter with the same glasses but he has to keep the letter 50cm away. What power of lens should he now use?

Answer

On the 50th birthday, he reads the card at a distance 25cm using a glass of +2.5D.Ten years later, his near point must have changed.
So after ten years,
$\text{u}=-50\text{cm},$
$\text{f}=\frac{1}{2.5\text{D}}=0.4\text{m}=40\text{cm}$
$\text{v}=\text{near point}$
Now, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow \frac{1}{\text{v}}=\frac{1}{\text{u}}+\frac{1}{\text{f}}=\frac{1}{-50}+\frac{1}{40}=\frac{1}{200}$
So, near point = v = 200cm
To read the farewell letter at a distance of 25cm,
U = –25cm
For lens formula,
$\Rightarrow\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{f}}=\frac{1}{200} -\frac{1}{-25}=\frac{1}{200}+\frac{1}{25}=\frac{9}{200}$
$\Rightarrow\text{f}=\frac{200}{9}\text{cm}=\frac{2}{9}\text{m}$
⇒ Power of the lens $=\frac{1}{\text{f}}=\frac{9}{2}=4.5\text{D.}$
$\therefore$ He has to use a lens of power +4.5D.

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Question 885 Marks

An equiconvex lens with radii of curvature of magnitude r each, is put over a liquid layer poured on top of a plane mirror. A small needle, with its tip on the principal axis of the lens, is moved along the axis until its inverted real image coincides with the needle itself. The distance of the needle from the lens is measured to be 'a'. On removing the liquid layer and repeating the experiment the distance is found to be 'b'.

Given that two values of distances measured represent the focal length values in the two cases, obtain a formula for the refractive index of the liquid.

If r = 10cm, a = 15cm, b = 10cm, find the refractive index of the liquid.

Answer

The focal length $(f_1)$ of lens is given by

$\frac{1}{\text{f}_1}=(\text{n}-1)(\frac{1}{\text{r}}+\frac{1}{\text{r}})=\frac{2(\text{n}-1)}{\text{r}}$
Given $f_1= b$
$\Rightarrow\frac{1}{\text{b}}=\frac{2(\text{n}-1)}{\text{r}}$
$\Rightarrow\text{b}=\frac{\text{r}}{2(\text{n}-1)}$
The focal lenght of liquid lens (plano concave lens) is
$\frac{1}{\text{f}_2}=(\text{n}_{\text{l}}-1)(-\frac{1}{\text{r}}-\frac{1}{\propto})$
$=-\frac{(\text{n}_{\text{l}}-1)}{\text{r}}$
$\Rightarrow\text{f}_2=\frac{\text{r}}{(\text{n}_{\text{l}}-1)}$
As glass lens and liquid lens are in contact
$\therefore\frac{1}{\text{f}}=\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}=\frac{1}{\text{b}}-\frac{(\text{n}_{\text{l}}-1)}{\text{r}}$
Given $f = a$
$\therefore\frac{1}{\text{a}}=\frac{1}{\text{b}}-\frac{\text{n}_{\text{l}}-1}{\text{r}}$
$\Rightarrow\frac{\text{n}_{\text{l}}-1}{\text{r}}=\frac{1}{\text{b}}-\frac{1}{\text{a}}$
$\Rightarrow\text{n}_{\text{l}}-1=\text{r}\big(\frac{1}{\text{b}}-\frac{1}{\text{a}}\big)$
Refractive index of liquid,

$\text{n}_{\text{l}}=1+\frac{\text{r}}{\text{b}}-\frac{\text{r}}{\text{a}}$

$\text{n}_{\text{l}}=1+\frac{10}{10}-\frac{10}{15}$
$=1+1-\frac{2}{3}=2-\frac{2}{3}=\frac{4}{3}=1.33$

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Question 895 Marks

In many experimental set-ups the source and screen are fixed at a distance say D and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.

Answer

According to prinicipal of reversibility, the position of object and image ate interchangeable. According to lens maker's formula, We have $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}\ .....(1)$ Hence, there are two positions for which there shall be an image. Let us suppose the first position be when the lens is at O. Finding u and v ans substituting in lens formula. According to the question, we have -u + v = D ⇒ u = -(D - v) Substituting u = -(D - v) in (1), we get $\frac{1}{\text{v}}+\frac{1}{\text{D}-\text{v}}=\frac{1}{\text{f}}$ $\Rightarrow\ \frac{\text{D}-\text{v}+\text{v}}{(\text{D}-\text{v})\text{v}}=\frac{1}{\text{f}}$ $\Rightarrow\ \text{v}^2-\text{Dv}+\text{Df}=0$ $\Rightarrow\ \text{v}=\frac{\text{D}}{2}\pm\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}$ $\therefore\ \text{u}=-(\text{D}-\text{v})=-\Big(=\frac{\text{D}}{2}\pm\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}\Big)$

When, the object distance is $\frac{\text{D}}{2}+\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}$, then the image forms at $=\frac{\text{D}}{2}-\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}$ Similarly, when the object distance is $=\frac{\text{D}}{2}-\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}$, then the image forms at $=\frac{\text{D}}{2}+\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}$ The distance between the poles for these two object distance is given by v - u $=\frac{\text{D}}{2}+\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}-\bigg(\frac{\text{D}}{2}-\frac{\sqrt{\text{D}^2-4\text{Df}}}{2}\bigg)$ $=\sqrt{\text{D}^2-4\text{Df}}$ Let us suppose $\text{d}=\sqrt{\text{D}^2-4\text{Df}}$ if $\text{u}=\frac{\text{D}}{2}+\frac{\text{d}}{2}$, then the image is at $\text{v}=\frac{\text{D}}{2}-\frac{\text{d}}{2}$,then the magnification is given by $\text{m}_1=\frac{\text{v}}{\text{u}}=\frac{\text{D}-\text{d}}{\text{D}+\text{d}}$ If $\text{u}=\frac{\text{D}-\text{d}}{2}$, then $\text{v}=\frac{\text{D}+\text{d}}{2}$, then the magnification is given by $\text{m}_2=\frac{\text{v}}{\text{u}}=\frac{\text{D}+\text{d}}{\text{D}-\text{d}}$ Thus, the ration of the images sizes is given by $\frac{\text{m}_2}{\text{m}_1}=\frac{\frac{\text{D}+\text{d}}{\text{D}-\text{d}}}{\frac{\text{D}+\text{d}}{\text{D}-\text{d}}}=\Big({\frac{\text{D}+\text{d}}{\text{D}-\text{d}}}\Big)^2$

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Question 905 Marks

Find the maximum magnifying power of a compound Microscope having a 25 diopter lens as the objective, a 5 diopter lens as the eyepiece and the separation 30cm between the two lenses. The least distance for clear vision is 25cm.

Answer

For the given compound microscope $\text{f}_0=\frac{1}{25\text{ diopter}}=0.04\text{m}=4\text{cm,}$ $\text{f}_\text{e}=\frac{1}{5\text{ diopter}}=0.2\text{m}=20\text{cm}$ D = 25cm,
separation between objective and eyepiece = 30cm
The magnifying power is maximuwm hen the image is formed by the eye piece at least distance of clear vision
i.e. D = 25cm for the eye piece, $v_e= -25cm, f_e = 20cm$
For lens formula, $\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{f}_\text{e}}$
$\Rightarrow\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{f}_\text{e}}$
$\Rightarrow\frac{1}{\text{u}_\text{e}}\frac{1}{-25}-\frac{1}{20} $
$\Rightarrow\text{u}_\text{e}=11.11\text{cm}$
So, for the objective lens, the image distance should be
$v_0 = 30 - (11.11) = 18.89$cm Now, for the objective lens,
$v_0 = +18.89$cm (because real image is produced) $f_0 = 4cm$
So, $\frac{1}{\text{u}_\text{o}}=\frac{1}{\text{v}_\text{o}}-\frac{1}{\text{f}_\text{o}}\Rightarrow \frac{1}{18.89}-\frac{1}{4}\\=0.053-0.25=-0.197$
$\Rightarrow\text{u}_\text{o}=-5.07\text{cm}$
So, the maximum magnificent power is given by
$\text{m}=-\frac{\text{v}_\text{o}}{\text{u}_\text{o}}\Big[1+\frac{\text{D}}{\text{f}_\text{e}}\Big]=\frac{18.89}{-5.07}\Big[1+\frac{25}{20}\Big]$$=3.7225\times2.25=8.376$

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Question 915 Marks

An infinitely long cylinder of radius R is made of an unusual exotic material with refractive index -1 (Fig). The cylinder is placed between two planes whose normals are along the y direction. The center of the cylinder O lies along the y-axis. A narrow laser beam is directed along the y direction from the lower plate. The laser source is at a horizontal distance x from the diameter in the y direction. Find the range of x such that light emitted from the lower plane does not reach the upper plane.

Answer

We are given that, the refractive index of the material is -1, $\theta_\text{r}$, is negative and $\theta_\text{r}'$ is positive.
Now, $|\theta_\text{i}|=|\theta_\text{r}|=|\theta_\text{r}'|$
The total deviation of the out coming ray from the incoming ray is $4\theta_\text{i}$.
Rays shall not reach the receiving plate if $\frac{\pi}{2}\leq4\theta,\leq\frac{3\pi}{2}$ [angles measured clockwise from the y-axis]
Dividing all the terms by 4, we get
$\frac{\pi}{8}\leq\theta,\leq\frac{3\pi}{8}\ .....(1)$

Now, $\sin\theta_\text{i}=\frac{\text{x}}{\text{R}}$
$\Rightarrow\ \theta_\text{i}=\sin^{-1}\Big(\frac{\text{x}}{\text{R}}\Big)$
Substituting $\theta_\text{i}=\sin^{-1}\Big(\frac{\text{x}}{\text{R}}\Big)$ in (1), we get
$\frac{\pi}{8}\leq\sin^{-1}\frac{\text{x}}{\text{R}}\leq\frac{3\pi}{8}$
$\frac{\pi}{8}\leq\frac{\text{x}}{\text{R}}\leq\frac{3\pi}{8}$
$\Rightarrow\ \frac{\pi\text{R}}{8}\leq\text{x}\leq\frac{3\pi\text{R}}{8}$
Thus, the range of x such that light emitted from the lower plane does not reach the upper plane is
$\frac{\text{R}\pi}{8}\leq\text{x}\leq\frac{\text{R}3\pi}{8}.$

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Question 925 Marks

Draw a ray diagram to show the formation of real image of the same size as that of the object placed in front of a converging lens. Using this ray diagram establish the relation between u, v and f for this lens.

Answer

Thin Lens Formula: Suppose an object AB of finite size is placed normally on the principal axis of a thin convex lens (fig.). A ray AP starting from A parallel to the principal axis, after refraction through the lens, passes through the second focus F. Another ray AC directed towards the optical centre C of the lens, goes straight undeviated. Both the rays meet at A′ Thus A′ is the real image of A. The perpendicular A′ B′ dropped from A′ on the principal axis is the whole image of AB.

Let distance of object AB from lens = u Distance of image A′B′ from lens = v Focal length of lens = f. We can see that Triangles ABC and A′B′C′ are similar $\frac{\text{AB}}{\text{A}'\text{B}'}=\frac{\text{CB}}{\text{C}'\text{B}'}\dots(\text{i})$ Similarly triangles PCF and A'B'F are similar $\frac{\text{P}\text{C}}{\text{A}'\text{B}'}=\frac{\text{CF}}{\text{F}'\text{B}}$ But PC = AB $\frac{\text{AB}}{\text{A}'\text{B}'}=\frac{\text{CF}}{\text{FB}'}\dots(\text{ii})$ From (i) and (ii), we get $\frac{\text{CB}}{\text{CB}'}=\frac{\text{CF}}{\text{FB}'}\dots(\text{iii})$ From sign convention, CB = -u, CB' = v, CF = f and FB' = CB' - CF = v - f Substituting this value in (iii), we get, $-\frac{\text{u}}{\text{v}}=\frac{\text{f}}{\text{v}-\text{f}}$ or -u (v - f) = vf or -uv + uf = uf Dividing throughout by uvf, we get $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\dots(\text{iv})$

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Question 935 Marks

If light passes near a massive object, the gravitational interaction causes a bending of the ray. This can be thought of as happening due to a change in the effective refrative index of the medium given by
$\text{n}(\text{r})=1+2\frac{\text{GM}}{\text{rc}^2}$
where r is the distance of the point of consideration from the centre of the mass of the massive body, G is the universal gravitational constant, M the mass of the body and c the speed of light in vacuum. Considering a spherical object find the deviation of the ray from the original path as it grazes the object.

Answer

Let us consider two spherical surfaces of radius r and r + dr. Let the light be incident at an angle $\theta$ at the surface at r and leave r + dr at an angle $\theta+\text{d}\theta$. Then from Snell's law,
$\text{n}(\text{r})\sin\theta=\text{n}(\text{r}+\text{dr})\sin(\theta+\text{d}\theta)$
$=\bigg(\text{n}(\text{r})+\Big(\frac{\text{dn}}{\text{dr}}\Big)\text{dr}\bigg)(\sin\theta.\cos\text{d}\theta+\cos\theta.\sin\text{d}\theta)$
$\Rightarrow\ \text{n}(\text{r})\sin\theta=\bigg(\text{n}(\text{r})+\Big(\frac{\text{dn}}{\text{dr}}\Big)\text{dr}\bigg)(\sin\theta+\cos\theta\text{d}\theta)$
For small angle, $\sin\text{d}\theta\approx\text{ and }\cos\text{d}\theta\approx1$
Ignoring the product of differentials
$\Rightarrow\ \text{n}(\text{r})\sin\theta=\text{n}(\text{r}).\sin\theta+\Big(\frac{\text{dn}}{\text{dr}}\Big)\text{dr}.\sin\theta+\text{n}(\text{r}).\cos\theta.\text{d}\theta$
or we have, $-\frac{\text{dn}}{\text{dr}}\tan\theta=\text{n}(\text{r})\frac{\text{d}\theta}{\text{dr}}$
$\frac{2\text{GM}}{\text{r}^2\text{c}^2}\tan\theta=\Big(1+\frac{2\text{GM}}{\text{rc}^2}\Big)\frac{\text{d}\theta}{\text{dr}}\approx\frac{\text{d}\theta}{\text{dr}}$
$\int_0^\theta\text{d}\theta=\frac{2\text{GM}}{\text{c}^2}\int_{-\infty}^{\infty}\frac{\tan\theta\text{dr}}{\text{r}^2}$
Now, $\text{r}^2=\text{x}^2+\text{R}^2\text{ and }\tan\theta=\frac{\text{R}}{\text{x}}$
$2\text{rdr}=2\text{xdx}$
Now subsituting for integtals, we have
$\int_0^\theta\text{d}\theta=\frac{2\text{GM}}{\text{c}^2}\int_{-\infty}^{\infty}\frac{\text{R}}{\text{x}}\frac{\text{xdx}}{(\text{x}^2+\text{R}^2)^\frac{3}{2}}$
Put $\text{x}=\text{R}\tan\phi$
$\text{dx}=\text{R}\sec^2\phi\text{d}\phi$
$\therefore\ \theta_0=\frac{2\text{GMR}}{\text{c}^2}\int_{\frac{-\text{x}}{2}}^{\frac{\text{x}}{2}}\frac{\text{R}\sec^2\phi\text{d}\phi}{\text{R}^2\sec^3\phi}$
$\theta_0=\frac{2\text{GM}}{\text{Rc}^2}\int_{\frac{-\text{x}}{2}}^{\frac{\text{x}}{2}}\cos\phi\text{d}\phi=\frac{4\text{GM}}{\text{Rc}^2}$
$\Rightarrow\ \theta_0=\frac{4\text{GM}}{\text{Rc}^2}$. This is the required proof.

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Question 945 Marks

A convex lens has a focal length of 10cm. Find the location and nature of the image if a point object is placed on the principal axis at a distance of:

9.8cm
10.2cm from the lens.

Answer

Given that, f = 10cm

When u = -9.5cm

$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{F}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{10}-\frac{1}{9.8}=\frac{-0.2}{98}$

$\Rightarrow\text{v}=-490\text{cm}$

So, $\text{m}=\frac{\text{v}}{\text{u}}=\frac{-490}{-9.8}=50\text{cm}$

So, the image is erect and virtual.

When u = -10.2cm

$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{F}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{10}-\frac{1}{-10.2}=\frac{102}{0.2}$

$\Rightarrow\text{v}=510\text{cm}$

So, $\text{m}=\frac{\text{v}}{\text{u}}=\frac{510}{-9.8}$

The image is real and inverted.

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Question 955 Marks

A 5mm high pin is placed at a distance of 15cm from a convex lens of focal length 10cm. A second lens of focal length 5cm is placed 40cm from the first lens and 55cm from the pin. Find

The position of the final image.
Its nature.
Its size.

Answer

First lens:

u = -15cm, f = 10cm
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}-\Big(-\frac{1}{15}\Big)=-\frac{1}{10}$
$\Rightarrow\text{v}=30\text{cm}$
So, the final image is formed 10cm right of second lens.

m for $1^{st}$ lens:

$\frac{\text{v}}{\text{u}}=\frac{\text{h}_{\text{image}}}{\text{h}_{\text{object}}}\Rightarrow\Big(\frac{30}{-15}\Big)=\frac{\text{h}_{\text{image}}}{5{\text{mm}}}$
$\Rightarrow\text{h}_{\text{image}}=-10\text{mm}$ (inverted)
Second lens:
$u = -(40 - 30) = -10cm; f = 5cm$
[since, the image of $1^{st}$ lens becomes the object for the second lens]
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}-\Big(-\frac{1}{10}\Big)=\frac{1}{5}$
$\Rightarrow\text{v}=10\text{cm}$
m for $2^{nd}$ lens:
$\frac{\text{v}}{\text{u}}=\frac{\text{h}_{\text{image}}}{\text{h}_{\text{object}}}\Rightarrow\Big(\frac{10}{10}\Big)=\frac{\text{h}_{\text{image}}}{{-10}}$
$\Rightarrow\text{h}_{\text{image}}=10\text{mm}$ (erect, real).

So, size of final image = 10mm

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Question 965 Marks

A myopic adult has a far point at 0.1m. His power of accomodation is 4 diopters.

What power lenses are required to see distant objects?
What is his near point without glasses?
What is his near point with glasses? (Take the image distance from the lens of the eye to the retina to be 2cm.)

Answer

If two thin lenses of focal lenght $f_1$ and $f_2$ are in contact, then the effective focal length of the combination is given by
$\frac{1}{\text{f}}=\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}$
Also the effective power of the combination is given by
$P = P_1 + P_2$

Let the power at the far point be $P_f$ for the normal relaxed eye of an average person.

Now, $\text{P}_\text{f}=\frac{1}{\text{f}}=\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}$
Substituting $f_1 = 0.1$ and $\text{f}_2=\frac{1}{40}=0.02$, we get
$\text{P}_\text{f}=\frac{1}{\text{f}}=\frac{1}{0.1}+\frac{1}{0.02}=60\text{D}$
By the corrective lens the object distance at the far point is infinity.
The power required is given by
$\text{P}_\text{f}=\frac{1}{\text{f}'}=\frac{1}{\infty}+\frac{1}{0.02}=50\text{D}$
For eye + lens system, we have the sum of the eye and that of the lgasses $P_g$
$\therefore\ \text{P}_\text{f}=\text{P}_\text{f}+\text{P}_\text{g}$
$\Rightarrow\ \text{P}_\text{g}=\text{P}_\text{f}-\text{P}_\text{f}$
Substituting $P_f = 60D$ and $P'_f = 50D$, we get
$\text{P}_\text{g}=-10\text{D}$

We are given that the power of accommodation of adult is 4D for the normal eye.

Let us suppose $P_n$ be the power of the normal eye for near vision.
$\therefore\ 4=\text{P}_\text{n}-\text{P}_\text{f}$
or $\text{P}_\text{n}=64\text{D}$
Let his near points be $x_n$, then
$\frac{1}{\text{x}_\text{n}}+\frac{1}{0.02}=64$
$\Rightarrow\ \frac{1}{\text{x}_\text{n}}+50=64$
$\Rightarrow\ \frac{1}{\text{x}_\text{n}}=14,$

Now, $P'_n = P'_f + 4 = 54$

Let his near point be $x'_n$, then
$54=\frac{1}{\text{x}'_\text{n}}+\frac{1}{0.02}$
$\Rightarrow\ 54=\frac{1}{\text{x}'_\text{n}}+50$
$\Rightarrow\ \frac{1}{\text{x}'_\text{n}}=4$
$\Rightarrow\ \text{x}'_\text{n}=\frac{1}{4}=0.25\text{m}$

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Question 975 Marks

A thin lens made of a material of refractive index $\mu_2$ has a medium of refractive index $\mu_1$ on one side and a medium of refractive index $\mu_3$ on the other side. The lens is biconvex and the two radii of curvature have equal magnitude R. A beam of light travelling parallel to the principal axis is incident on the lens. Where will the image be formed if the beam is incident from:

The medium $\mu_1$
From the medium $\mu_3?$

Answer

When the beam is incident on the lens from medium $\mu_{3}$

Then $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$ or $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{(-\infty)}=\frac{\mu_2-\mu_1}{\text{R}}$
or $\frac{1}{\text{v}}=\frac{\mu_2-\mu_1}{\mu_2\text{R}}$ or $\text{v}=\frac{\mu_2\text{R}}{\mu_2-\mu_1}$
Again, for $2^{nd} $ refraction, $\frac{\mu_3}{\text{v}}-\frac{\mu_2}{\text{u}}=\frac{\mu_3-\mu_2}{\text{R}}$
or, $\frac{\mu_3}{\text{v}}=-\Big[\frac{\mu_3-\mu_2}{\text{R}}-\frac{\mu_2}{\mu_2\text{R}}(\mu_2-\mu_1)\Big]\Rightarrow-\Big[\frac{\mu_3-\mu_2-\mu_2+\mu_1}{\text{R}}\Big]$
or, $\text{v}=-\Big[\frac{\mu_3\text{R}}{\mu_3-2\mu_2+\mu_1}\Big]$
So, the image will be formed at $=\frac{\mu_3\text{R}}{2\mu_2-\mu_1-\mu_3}$

Similarly for the beam from $\mu_3$ medium the image is formed at $\frac{\mu_3\text{R}}{2\mu_2-\mu_1-\mu_3}$

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Question 985 Marks

Consider the situation in figure. The bottom of the pot is a reflecting plane mirror, S is a small fish and T is a human eye. Refractive index of water is.

At what distance(s) from itself will the fish see the image(s) of the eye?
At what distance(s) from itself will the eye see the image(s) of the fish.

Answer

Let x = distance of the image of the eye formed above the surface as seen by the fish

So, $\frac{\text{H}}{\text{x}}=\frac{\text{Real depth}}{\text{Apparent depth}}=\frac{1}{\mu}$ or $\text{x}=\mu\text{H}$

So, distance of the direct image $=\frac{\text{H}}{2}+\mu\text{H}=\text{H}\Big(\mu+\frac{1}{2}\Big)$

Similarly, image through mirror $=\frac{\text{H}}{2}+(\text{H}\times\text{x})=\frac{3\text{H}}{2}+\mu\text{H}=\text{H}\Big(\mu+\frac{3}{2}\Big)$

Here, $\frac{\frac{\text{H}}{2}}{\text{y}}=\mu,$ So, $\text{y}=\frac{\text{H}}{2\mu}$

Where, y = distance of the image of fish below the surface as seen by eye.

So, Direct image $=\text{H}+\text{y}=\text{H}+\frac{\text{H}}{2\mu}=\text{H}\Big(1+\frac{1}{2\mu}\Big)$

Again another image of fish will be formed $\frac{\text{H}}{2}$ below the mirror.

So, the real depth for that image of fish becomes $\text{H}+\frac{\text{H}}{2}=\frac{3\text{H}}{2}$

So, Apparent depth from the surface of water $=\frac{3\text{H}}{2\mu}$

So, distance of the image from the eye $=\text{H}+\frac{3\text{H}}{2\mu}=\text{H}\Big(1+\frac{3}{2\mu}\Big).$

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Question 995 Marks

An object is placed at a distance of 30cm from a converging lens of focal length 15cm. A normal eye (near point 25cm, far point infinity) is placed close to the lens on the other side.

Can the eye see the object clearly?
What should be the minimum separation between the lens and the eye so that the eye can clearly see the object?
Can a diverging lens, placed in contact with the converging lens, help in seeing the object clearly when the eye is close to the lens?

Answer

Object distance, u = -30cm Focal length, f = 15cm Image distance, v = ? The lens formula is given by$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{v}}-\frac{1}{-30}=\frac{1}{15}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{30}$
$\Rightarrow \text{v}=+30\text{cm}$
(on the opposite side of the object)

No, the eye placed close to the lens cannot see the object clearly.
The eye should be 30cm away from the lens to see the object clearly.
The diverging lens will always form an image at a large distance from the eye this image cannot be seen through the human eye.

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Question 1005 Marks

Three immiscible liquids of densities $d_1 > d_2 > d_3$ and refractive indices $\mu _1 > \mu _2 > \mu _3$ are put in a beaker. The height of each liquid column is $\frac{\text{h}}{3}$. A dot is made at the bottom of the beaker. For near normal vision, find the apparent depth of the dot.

Answer

Key concept: Coordinate convention: At the first surfece (+upward and -ve downward) $\frac{\mu_2}{\text{h}'}-\frac{\mu_1}{(-\text{h})}=\frac{\mu_2-\mu_1}{\infty}$ (infinity because the surface is plane), or $\text{h}'=-\frac{\mu_2}{\mu_1}\text{h}$. The negative sign shown that it is on the side of the object h' is the apparent depth of O after refraction from interface.

The position of image if O after refraction from surface-1. If seen from $\mu_2$ the apparent depth is $ h_1. \text{h}_1=-\frac{\mu_2}{\mu_1}\frac{\text{h}}{3}$ The nagative sign shows that it is on the side of the object.

Since, the image formed by surface-1 will act as an object for surface-2. IF seen form $\mu_3$, the apparent depth is $h_2$. Similarly, the image formed by Medium 2, $O_2$ acts as an object for Medium 3. $\text{h}_2=-\frac{\mu_3}{\mu_2}\Big(\frac{\mu_2}{\mu_1}\frac{\text{h}}{3}+\frac{\text{h}}{3}\Big)=-\frac{\text{h}}{3}\Big(\frac{\mu_3}{\mu_1}+\frac{\mu_2}{\mu_1}\Big)$ Finally the oimage formed by surface-2 will act as an object for surface-2. If seen from outside, the apparent depth is $h_3. \text{h}_3=-\frac{1}{\mu_3}\bigg[\frac{\text{h}}{3}+\frac{\text{h}}{3}\Big(\frac{\mu_3}{\mu_2}+\frac{\mu_3}{\mu_1}\Big)\bigg]=-\frac{\text{h}}{3}\Big(\frac{1}{\mu_1}+\frac{1}{\mu_2}+\frac{1}{\mu_3}\Big)$ Hence apparent depth of dot is $\frac{\text{h}}{3}\Big(\frac{1}{\mu_1}+\frac{1}{\mu_2}+\frac{1}{\mu_3}\Big)$.Important point:
Apparent depth (distance of final image from final surface) $=\frac{\text{t}_1}{\text{n}_{1\text{real}}}+\frac{\text{t}_2}{\text{n}_{2\text{real}}}+\frac{\text{t}_3}{\text{n}_{3\text{real}}}+\ .....\ +\frac{\text{t}_\text{n}}{\text{n}_\text{nreal}}$

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