2 Marks Questions - Physics STD 12 Science Questions - Vidyadip

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2 Marks Questions

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Question 12 Marks

Define the magnifying power of a compound microscope when the final image is formed at infinity. Why must both the objective and the eyepiece of a compound microscope has short focal lengths? Explain.

Answer

When the final image is formed at infinity.
When the final image is formed at infinity, the angular magnification due to the eyepiece is
$\text{m}_e=D/\text{f}_e$
Thus, the total magnification when the image is formed at infinity can be defined as the product of magnification of objective lens and eyepiece. i.e
$m=m_0 m_e=\left(L / f_0\right)\left(D / f_e\right)$
From the above the equation, we can see that to achieve a large magnification of a small object, the objective and eyepiece should have small focal lengths.

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Question 22 Marks

Answer the following question:
Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?

Answer

The focal length of a convex lens cannot be decreased by a greater amount. This is because making lenses having very small focal lengths is not easy. Spherical and chromatic aberrations are produced by a convex lens having a very small focal length.

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Question 32 Marks

Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision?

Answer

No, a person may have normal ability of accommodation yet, he may be myopic or hypermetropic.
In fact, myopia arises when length of eye ball (from front to back) gets elongated and hypermetropia arises when length of eye ball gets shortened.
However, when eye ball has normal length, but the eye-lens losses partially its power of accommodation, this defect is called presbiopia.

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Question 42 Marks

A small pin fixed on a table top is viewed from above from a distance of 50cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?

Answer

Actual depth of the pin, d= 15 cm
Apparent dept of the pin = d'
Refractive index of glass, µ = 1.5
Ratio of actual depth to the apparent depth is equal to the refractive index of glass, i.e.
$\mu=\frac{\text{d}}{\text{d}'}$
$\therefore \ \text{d}'=\frac{\text{d}}{\mu}$
$=\frac{15}{1.5}=10 \ \text{cm}$
The distance at which the pin appears to be raised = d'− d
= 15 - 10 = 5 cm
For a small angle of incidence, this distance does not depend upon the location of the slab.

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Question 52 Marks

A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.

Answer

Distance between the image (screen) and the object, D = 90 cm
Distance between two locations of the convex lens, d = 20 cm
Focal length of the lens = f
Focal length is related to d and D as:
$\text{f}=\frac{\text{D}^2-\text{d}^2}{4\text{D}}$
$=\frac{(90)^2-(20)^2}{4\times90}=\frac{7700}{360}=\frac{700}{36}=21.39 \ \text{cm}$

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Question 62 Marks

Answer the following question:
The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter?

Answer

Yes
The refractive index of diamond (2.42) is more than that of ordinary glass (1.5). The critical angle for diamond is less than that for glass. A diamond cutter uses a large angle of incidence to ensure that the light entering the diamond is totally reflected from its faces. This is the reason for the sparkling effect of a diamond.

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Question 72 Marks

Answer the following question:
virtual image, we always say, cannot be caught on a screen.
Yet when we 'see' a virtual image, we are obviously bringing it on to the 'screen' (i.e., the retina) of our eye. Is there a contradiction?

Answer

No
A virtual image is formed when light rays diverge. The convex lens of the eye causes these divergent rays to converge at the retina. In this case, the virtual image serves as an object for the lens to produce a real image.

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Question 82 Marks

Answer the following question:
When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?

Answer

When we place our eyes too close to the eyepiece of a compound microscope, we are unable to collect much refracted light. As a result, the field of view decreases substantially. Hence, the clarity of the image gets blurred.
The best position of the eye for viewing through a compound microscope is at the eye-ring attached to the eyepiece. The precise location of the eye depends on the separation between the objective lens and the eyepiece.

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Question 92 Marks

The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

Answer

Distance between the object and the image, $d = 3\ m$
Maximum focal length of the convex lens $= f_{max}$
For real images, the maximum focal length is given as:
$\text{f}_\text{max}=\frac{\text{d}}{4}$
$=\frac{3}{4}=0.75\text{m}$
Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.

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Question 102 Marks

Answer the following question:
The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?

Answer

Though the image size is bigger than the object, the angular size of the image is equal to the angular size of the object. A magnifying glass helps one see the objects placed closer than the least distance of distinct vision (i.e., 25 cm). A closer object causes a larger angular size. A magnifying glass provides angular magnification. Without magnification, the object cannot be placed closer to the eye. With magnification, the object can be placed much closer to the eye.

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Question 112 Marks

Answer the following question:
You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.

Answer

Yes, Plane and convex mirrors can produce real images as well. If the object is virtual, i.e., if the light rays converging at a point behind a plane mirror (or a convex mirror) are reflected to a point on a screen placed in front of the mirror, then a real image will be formed.

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Question 122 Marks

Answer the following question:
Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease?

Answer

Yes;
Decrease The apparent depth of a tank of water changes when viewed obliquely. This is because light bends on travelling from one medium to another. The apparent depth of the tank when viewed obliquely is less than the near-normal viewing.

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Question 132 Marks

Answer the following question:
A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is?

Answer

The diver is in the water and the fisherman is on land (i.e., in air). Water is a denser medium than air. It is given that the diver is viewing the fisherman. This indicates that the light rays are travelling from a denser medium to a rarer medium. Hence, the refracted rays will move away from the normal. As a result, the fisherman will appear to be taller.

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Question 142 Marks

Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards. A current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

Answer

Angle of deflection, θ = 3.5°
Distance of the screen from the mirror, D = 1.5 m
The reflected rays get deflected by an amount twice the angle of deflection i.e., 2θ = 7.0°
The displacement (d) of the reflected spot of light on the screen is given as:
$\tan2\theta=\frac{\text{d}}{1.5}$
$\therefore \ \text{d}=1.5\times\tan7^\circ=0.814 \ \text{m}=18.4 \ \text{cm}$
Hence, the displacement of the reflected spot of light is 18.4 cm.

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Question 152 Marks

Answer the following question:
In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?

Answer

Yes, the angular magnification changes. When the distance between the eye and a magnifying glass is increased, the angular magnification decreases a little. This is because the angle subtended at the eye is slightly less than the angle subtended at the lens. Image distance does not have any effect on angular magnification.

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Question 162 Marks

Is the magnification equal to the magnifying power in this case? Explain.

Answer

Near point, D = 25 cm
Distance where the object is kept, u = 7.14 cm
Magnifying power of the lens, $\text{m}=\frac{\text{D}}{\text{u}}$
$=\frac{25}{7.14}$
= 3.5
Here, in this case since the image is formed at the least distance of distinct vision, the magnifying power is equal to the magnitude of magnification.

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Question 172 Marks

A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?

Answer

The defect is called Astigmatism. When the curvature of the cornea plus eye-lens refracting system is not the same in different planes, this defect arises. As vertical lines are seen distinctly, the curvature in the vertical plane is enough, but the curvature of the lens system is insufficient in the horizontal plane and, the lines cannot be seen distinctly.
This defect is removed by using a cylindrical lens with its axis along the vertical plane.

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Question 182 Marks

Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled prism ABC. The refractive indices of the glass prism for the two rays ‘1’ and ‘2’ are respectively 1.38 and 1.52. Trace the path of these rays after entering through the prism.

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Question 192 Marks

A ray PQ incident normally on the refracting face BA is refracting in the prism BAC made of material of refractive index 1.5. Complete the path of ray through the prism. From which face will the ray emerge? Justify your answer.

Answer

Face-AC
Here $\text{i}_{c} =\sin^{-1}(\frac{2}{3})$
$ = \sin^{-1}(0.6)$
$\angle\text{i}$ on face AC is 30º which is less than $\angle\text{i}_{c}$ Hence the ray get replaced here.

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Question 202 Marks

Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled prism ABC. The refractive indices of the glass prism for the two rays ‘1’ and ‘2’ are respectively 1.3 and 1.5 Trace the path of these rays after entering through the prism.

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Question 212 Marks

A ray PQ incident normally on the refracting face BA is refracting in the prism BAC made of material of refractive index1.5. Complete the path of ray through the prism. From which face will the ray emerge? Justify your answer.

Answer

Face-AC
Here $\text{i}_{c} =\sin^{-1}(\frac{2}{3})$
$ = \sin^{-1}(0.6)$
$\angle\text{i}$ on face AC is 30º which is less than $\angle\text{i}_{c}$ Hence the ray get replaced here.

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Question 222 Marks

Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled prism ABC. The refractive indices of the glass prism for the two rays ‘1’ and ‘2’ are respectively 1.35 and 1.45. Trace the path of these rays after entering through the prism.

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Question 232 Marks

A convex lens of focal length $f_1$ is kept in contact with a concave lens of focal length $f_2$. Find the focal length of the combination.

Answer

For convex lens of focal length $(+ f_1)$
$ + \frac{1}{\text{f}_{1}} = \frac{1}{\text{v}'} - \frac{1}{\text{u}}$ - - - - - - - -(1)
For concave lens of focal length $(– f_2)$
$-\frac{1}{\text{f}_{2}} = \frac{1}{\text{v}} - \frac{1}{\text{v}'}$ - - - - - - - - (2)
Adding equation (1) and (2)
$\frac{1}{\text{f}_{1}} - \frac{1}{\text{f}_{2}} = \frac{1}{\text{v}} - \frac{1}{\text{u}}$ - - - - - - - - - -(3)
For an equivalent lens (using lens formula)
$\frac{1}{\text{f}} = \frac{1}{v} - \frac{1}{u}$ where f is the focal length of combination. - - - - - - - - - (4)
From equation (3) and (4),
$\frac{1}{\text{f}} =\frac{1}{\text{f}_{1}} - \frac{1}{\text{f}_{2}}.$

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Question 242 Marks

A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.

Answer

Capacitance with dielectric of thickness ‘t’
$\text{C} = \frac{\varepsilon_{0}\text{A}}{\text{d - t} + \frac{\text{t}}{\text{K}}}$
Put $\text{t} = \frac{\text{d}}{2}$
$\text{C} = \frac{\varepsilon_{0}\text{A}}{\text{d} - \frac{\text{d}}{2} + \frac{\text{d}}{2\text{K}}}$
$ = \frac{\varepsilon_{0}\text{A}}{\frac{\text{d}}{2} + \frac{\text{d}}{2\text{K}}}$
$ = \frac{\varepsilon_{0}\text{A}}{\frac{\text{d}}{2}\bigg(1+\frac{1}{\text{K}}\bigg)}$
$ = \frac{2\varepsilon_{0}\text{AK}}{\text{d}\text{(K} + 1)}.$

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Question 252 Marks

Describe briefly with the help of a circuit diagram, how the flow of current carriers in a p-n-p transistor is regulated with emitter-base junction forward biased and base-collector junction reverse biased.

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Question 262 Marks

Draw a labelled ray diagram of a reflecting telescope. Mention its two advantages over the refracting telescope.

Answer

Newtonion telescope in lieu of Cassegranian telescope:

Image formed is brighter.
Image is free from chromatic aberration.
Image is free from spherical aberration.
Higher resolving power.
Mechanical advantage in mounting.

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Question 272 Marks

Define refractive index of a transparent medium.
A ray of light passes through a triangular prism. Plot a graph showing the variation of the angle of deviation with the angle of incidence.

Answer

Refractive index: It is the ratio of speed of light in vacuum to the speed of light in medium.
Alternate Answer
$\mu= \frac{\sin\text{i}}{\sin\text{r}}$

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Question 282 Marks

Answer the following questions:

Optical and radio telescopes are built on the ground while X-ray astronomy is possible only from satellites orbiting the Earth. Why?
The small ozone layer on top of the stratosphere is crucial for human survival. Why ?

Answer

Atmosphere absorbs X rays, while visible and $\frac{\phi}{\text{I}} = \frac{\text{NAB}}{\text{k}}$ radiowaves can penetrate it.
Ozone layer absorbs ultraviolet radiations (harmful radiations) from sun and prevents it from reaching the earth’s surface and causing damage to life.

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Question 292 Marks

Draw a labelled ray diagram of an astronomical telescope in the near point position.
Write the expression for its magnifying power.

Answer

Magnifying Power:
$\text{m} = \frac{\text{f}_{0}}{\text{f}_{e}}\bigg(1 + \frac{\text{f}_{e}}{\text{D}}\bigg)\text{ or }\text{m} = \frac{\text{f}_{0}}{\text{f}_{e}}.$

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Question 302 Marks

Define resolving power of a compound microscope. How does the resolving power of a compound microscope change when.

refractive index of the medium between the object and objective lens increases?
wavelength of the radiation used is increased?

Answer

R.P. of microscope is reciprocal of the minimum separation of two near points, seen as distinct.
or
R.P. of microscope is reciprocal of the smallest distance between two point object, which are just resolved by the microscope.

increases, (with increase of $\mu$)
decreases, (with increase of $\lambda$).

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Question 312 Marks

Draw a labelled ray diagram to show the image formation in a refracting type astronomical telescope. Why should the diameter of the objective of a telescope be large?

Answer

Alternate Answer
High resolving power/Large light gathering power/Intensity of image more.

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Question 322 Marks

State the condition under which a large magnification can be achieved in an astronomical telescope.
Give two reasons to explain why a reflecting telescope is preferred over a refracting telescope.

Answer

$m=-\frac{f_0}{f_e}$

By increasing $?_0$ / decreasing $?_?$

No chromatic aberration.
No spherical aberration.
Mechanical advantage – low weight, easier to support.
Mirrors are easy to prepare.
More economical.

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Question 332 Marks

Use the mirror equation to show that an object placed between f and 2f of a concave mirror forms an image beyond 2f.

Answer

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\text{ }\text{ }\text{ }\text{ }\text{ }(f\text{ }\text{ } \text{is negative})$
$\text{U}=-\text{f}\Longrightarrow\frac{1}{v}=0\Longrightarrow v=\infty$
$\text{U}=-\text{2f}\Longrightarrow\frac{1}{v}=\frac{-1}{2f}\Longrightarrow v=-2f$
$\text{Hence if}\text{ }\text{ }\text{ }-\text{2f}<\text{u}<-\text{f}\Longrightarrow-2f
Alternate Answer
$2f>u>f$
$-\frac{1}{2f}>-\frac{1}{u}>-\frac{1}{f}$
$\frac{1}{f}-\frac{1}{2f}>\frac{1}{f}-\frac{1}{u}>\frac{1}{f}-\frac{1}{f}$
$\frac{1}{2f}<\frac{1}{V}<0$
$2\text{f}<\text{V}<\propto]$

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Question 342 Marks

Use the mirror equation to show that an object placed between f and 2f of a concave mirror produces a real image beyond 2f.

Answer

$\frac{1}{\text{f}} = \frac{1}{\text{v}}+ \frac{1}{\text{u}}$
For concave mirror f < 0 and u < 0 As object lies between f and 2f.
At u = – f
$\frac{1}{\text{v}} = - \frac{1}{\text{f}} +\frac{1}{\text{f}}$
$= > \text{v} = \propto$
At u = – 2f
$= > \frac{\text{1}}{\text{v}} = - \frac{\text{1}}{\text{f}} + \frac{\text{1}}{2\text{f}} = - \frac{\text{1}}{2\text{f}}$
$= > \text{v} =- 2\text{f}$
= > Hence, image distance v > – 2f
Since v is negative, therefore, the image is real.
Alternate Answer
$\frac{1}{\text{f}} =\frac{1}{\text{v}} +\frac{1}{\text{u}}$
For concave mirror
f < 0, u < 0
$\because2 \text{f}< \text{u}< \text{f}$
$\Rightarrow\frac{1}{2\text{f}}> \frac{1}{\text{u}}> \frac{1}{\text{f}}$
$\frac{1}{2\text{f}} -\frac{1}{\text{f}}>\frac{1}{\text{u}} -\frac{1}{\text{f}}> \frac{1}{\text{f}} - \frac{1}{\text{f}}$
$\Rightarrow - \frac{1}{2\text{f}} =\frac{1}{\text{v}}> 0 $ $\because\frac{1}{\text{u}} -\frac{1}{\text{f}} - \frac{1}{\text{-v}}$
$\Rightarrow\frac{1}{2\text{f}}<\frac{1}{\text{v}}< 0$
$\Rightarrow\text{v}< 0$
$\because$ Image is real
Also v > 2f image is formed beyond 2f.

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Question 352 Marks

Find an expression for intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids. In which position of the polaroid sheet will the transmitted intensity be maximum?

Answer

Let the rotating polaroid sheet makes an angle $\theta$with the first polaroid $\therefore$ Angle with the other polaroid will be $(90 - \theta)$

Applying Malus's law between $P_1$ and $P_3 $ $\text{I}' = \text{I}_{0}\cos^{2}\theta$ Between $P_3$ and $P_2$ $\text{I}'' = (\text{I}_{0}\cos^{2}\theta)\cos^{2}(90 - \theta)$ $\text{I}'' = \frac{I_{0}}{4}.\sin^{2}2\theta$ $\therefore$ Transmitted intensity will be maximum when $\theta = \frac{\pi}{4}$.

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Question 362 Marks

Why should the objective of a telescope have large focal length and large aperture? Justify your answer.

Answer

Large focal length: To increase magnifying power.
$\bigg(\because{m}=\frac{f_0}{f_e}\bigg)$
Large aperature: To increase resolving power.
$\bigg(\because{\text{RP}}=\frac{\text2a}{1.22\lambda}\bigg)$

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Question 372 Marks

A convex lens of focal length 25 cm is placed coaxially in contact with a concave lens of focal length 20 cm. Determine the power of the combination. Will the system be converging or diverging in nature?

Answer

Power of convex lens,
Power of concave lens,
Power of the combination $P=P_1+P_2=-1 D$
Nature: Diverging.

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Question 382 Marks

Write the necessary conditions for the phenomenon of total internal reflection to occur.
Write the relation between the refractive index and critical angle for a given pair of optical media.

Answer

1. Ray of light should travel from denser to rarer medium.
2. Angle of incidence shoud be more than the critical angle.
$\mu = \frac{1}{\sin\text{i}_{c}}$ where $i_c$ is the critical angle.

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Question 392 Marks

A ray of light, incident on an equilateral glass prism$(\mu_{g} = \sqrt{3})$ moves parallel to the base line of the prism inside it. Find the angle of incidence for this ray.

Answer

Fromthe diagram, $r = 30^\circ$
Also $\text{n}_{21} = \frac{\sin\text{i}}{\sin\text{r}}$
$\Rightarrow\sqrt{3} = \frac{\sin\text{i}}{\sin30}$
$\Rightarrow\sin\text{i} = \sqrt{3}\times\frac{1}{2}$
$\Rightarrow \text{i} = 60^{o}.$

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Question 402 Marks

Two convex lenses of same focal length but of aperture $A_1$ and $A_2\left(A_2<A_1\right)$, are used as the objective lenses in two astronomical telescopes having identical eyepieces. What is the ratio of their resolving power? Which telescope will you prefer and why? Give reason.

Answer

Ratio of resolving power $ = \frac{\text{A}_{1}}{\text{A}_{2}}$
Telescope of aperture $A_1$ is preferred.
Reason:
Higher resolving power/More light gathering power.

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Question 412 Marks

The bluish colour predominates in clear sky.
Violet colour is seen at the bottom of the spectrum when white light is dispersed by a prism.

State reasons to explain these observations.

Answer

As per Rayleigh’s law (scattering α 1/λ4), lights of shorter wavelengths scattered more by the atmospheric particles. This results in a dominance of bluish colour in the scattered light.
In the visible spectrum, violet light having its shortest wavelength, has the highest refractive index. Hence it is deviated the most.

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Question 422 Marks

The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. If focal length of the lens is 12 cm, find the refractive index of the material of the lens.

Answer

$\frac{1}{\text{f}} = (\mu - 1 )\bigg(\frac{1}{\text{R}_{1}} - \frac{1}{\text{R}_{2}}\bigg)$
$\frac{1}{12} = (\mu - 1 ) \bigg(\frac{1}{10} - \frac{1}{-15}\bigg)\Rightarrow\mu =1.5.$

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Question 432 Marks

A ray of light passing through an equilateral triangular glass prism from air undergoes minimum deviation when angle of incidence is 3/4th of the angle of the prism. Calculate the speed of light in the prism.

Answer

(HOST)$\text{D}_{m} = 2i - \text{A}; \text{D}_{m} = 2 \times \frac{3}{4}\text{A} - \text{A} = \text{A}/2$
$\therefore \text{D}_{m} = \frac{60^\circ}{2} = 30^{\circ}$
$\text{n} = \frac{c_1}{c_2} = \frac{\text{sin}\bigg(\frac{\text{A+D}_m}{2}\bigg)}{\text{sing}\frac{A}{2}}$
$\therefore \text{n} = \frac{c_1}{c_2} = \frac{\text{sin}45^\circ}{\text{sin}30^{\circ}} = \sqrt{2}$
$\text{c}_{2} = \frac{c_1}{\sqrt{2}} = \frac{3\times10^{8}}{\sqrt{}2} \text{ms}^{-1} = 1.5 \sqrt{2}\text{ms}^{-1}$
$\approx2 .12\times10^{8} \text{ms}^{-1}$
Alternate Answer
$\text{r}_{1} + \text{r}_{2} = \text{A}$
At minimum deviation,
$2\text{r} = \text{A}$
or $\text{r} = \frac{A}{2}$

$=30^{\circ}$
Also $\text{i} = \frac{3}{4}\times\text{A} = 45^{\circ}$
$\therefore \text{n} = \frac{c}{c_\circ} = \frac{sin\text{i}}{sin\text{r}}$
$\frac{sin45^\circ}{sin30^\circ} =\sqrt{2}$
$\text{c}_{2} = \frac{\text{c}_{1}}{\sqrt{2}} = \frac{3\times10^{8}\text{ms}^{-1}}{\sqrt{2}}$
$ = 1.5\sqrt{2}\times10^{8}\text{ms}^{-1}$
$\cong2.12 \times10^{8}\text{ms}^{-1}$

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Question 442 Marks

Draw a ray diagram of a reflecting type telescope. State two advantages of this telescope over a refracting telescope.

Answer

Advantages:

No chromatic aberration.
No spherical aberration.

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Question 452 Marks

A convex lens of refractive index 1.5 has a focal length of 18cm in air. Calculate the change in its focal length when it is immersed in water of refractive index $\frac{4}{3}$

Answer

$\frac{1}{\text{f}} = \big(\mu - 1\big)\bigg(\frac{1}{\text{R}_{1}} - \frac{1}{\text{R}_{2}}\bigg)$for air
$\frac{1}{\text{f}}_{a} = \big(^{a}\mu_{g} - 1\big)\bigg(\frac{1}{\text{R}_{1}} - \frac{1}{\text{R}_{2}}\bigg) = \bigg(\frac{1}{2}\bigg)\bigg(\frac{1}{\text{R}_{1}} - \frac{1}{\text{R}_{2}}\bigg)$
For water
$\frac{1}{\text{f}}_{w} = \big(^{w}\mu_{g} - 1\big)\bigg(\frac{1}{\text{R}_{1}} - \frac{1}{\text{R}_{2}}\bigg) = \bigg(\frac{1}{8}\bigg)\bigg(\frac{1}{\text{R}_{1}} - \frac{1}{\text{R}_{2}}\bigg)$
$\text{f}_{w} = 4 \text{f}_{a}$
Change in focal length $= 3 \text{f}_{a} = 54 \text{cm}$

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Question 462 Marks

Draw a labelled ray diagram of a reflecting type telescope. Write its any one advantage over refracting type telescope.

Answer

Advantage:No chromatic aberration/More light gathering power/Large size mirror can be more easily obtained than the lens.

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Question 472 Marks

Using lens maker’s formula, derive the thin lens formula $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$ for a biconvex lens.

Answer

Ace to lens maker's formula,
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=(\text{n}_{21}-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)\ ...(1)$
When object is at placed at infinity,
$\text{u}=\infty$
Image is obtained at focus,
$\text{v}=\text{f}$
Using these values in Eq (1),
$\frac{1}{\text{f}}-\frac{1}{\infty}={(\text{n}_{21}-1)}\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$
$\Rightarrow\frac{1}{\text{f}}=(\text{n}_{21}-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)\ ...(2)$
$\therefore$ By Eq (1) and (2),
$\Rightarrow\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$

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Question 482 Marks

The space between the plates of a parallel plate capacitor is completely filled in two ways. In the first case, it is filled with a slab of dielectric constant K. In the second case, it is filled with two slabs of equal thickness and dielectric constants $\mathrm{K}_1$ and $\mathrm{K}_2$ respectively as shown in the figure. The capacitance of the capacitor is same in the two cases. Obtain the relationship between $\mathrm{K}, \mathrm{K}_1$ and $\mathrm{K}_2$.

Answer

$\text{C}_1=\frac{\text{K}\varepsilon_0\text{A}}{\text{d}}$
$C_2=$ parallel combination of two capacitors,
$=\frac{\text{K}_1\varepsilon_0\big(\frac{\text{A}}{2}\big)}{\text{d}}+\frac{\text{K}_2\varepsilon_0\big(\frac{\text{A}}{2}\big)}{\text{d}}$
$\frac{\varepsilon_0\text{A}}{2\text{d}}\big(\text{K}_1+\text{K}_2\big)$
$\because\text{C}_1=\text{C}_2$
$\Rightarrow\text{K}=\frac{\text{K}_1+\text{K}_2}{2}$

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Question 492 Marks

N small conducting liquid droplets, each of radius r, are charged to a potential V each. These droplets coalesce to form a single large drop without any charge leakage. Find the potential of the large drop.

Answer

$\text{q}_{\text{new}}=\text{Nq};$ q = charge on each small droplet,
$\frac{4}{3}\pi\text{R}^3=\text{N}\Big(\frac{4}{3}\pi\text{R}^3\Big)$
$\Rightarrow\text{R}=\text{N}^{\frac{1}{3}}\text{r}$
R = radius of larger drop,
$\because\text{V}=\frac{\text{kq}}{\text{r}}=$ potential on each small dropper.
$\therefore$ V '= Potential on large drop,
$=\frac{\text{kq}_{\text{new}}}{\text{R}}=\frac{\text{K}(\text{Nq})}{\text{N}^{\frac{1}{3}}\text{r}}=\text{N}^\frac{2}{3}\Big(\frac{\text{kq}}{\text{r}}\Big)$
$\Rightarrow\text{V}'=\text{N}^{\frac{2}{3}}\text{V}$

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Question 502 Marks

Redraw the diagram given below and mark the position of the centre of curvature of the spherical mirror used in the given set up.

Answer

If the object is in between focus 'F' and centre of curvature 'C', image would be beyond the centre of curvature, inverted real and magnified.

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Question 512 Marks

Can a plane mirror ever form a real image?

Answer

It object is virtual image is real.

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Question 522 Marks

A 1cm object is placed perpendicular to the principal axis of a convex mirror of focal length 7.5cm. Find its distance from the mirror if the image formed is 0.6cm in size.

Answer

$\text{m}=-\frac{\text{v}}{\text{u}}=0.6$ and $\text{f}=7.5\text{cm}=\frac{15}{2}\text{cm}$
From mirror equation,
$\Rightarrow\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{0.6\text{u}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow \frac{5}{3\text{u}}-\frac{1}{\text{u}}=\frac{5}{15}$
$\Rightarrow\text{u}=5\text{cm}$

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Question 532 Marks

A concave lens is placed in water. Will there be any change in focal length? Give reason.

Answer

Focal length of lens in water $\text{f}_{\text{w}}=\frac{\text{n}_{\text{g}}-1}{\frac{\text{n}_{\text{g}}-1}{\text{nw}}-1}$
$\text{As n}_{\text{g}}>\text{n}_{\text{w}},\frac{\text{n}_{\text{g}}}{\text{n}_{\text{w}}}>1,\text{ so }\text{f}_{\text{w}}>\text{f}_{\text{a}}$

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Question 542 Marks

Can virtual image be formed on the retina in a seeing process?

Answer

The retina acts as a screen; only real images can be obtained on the screen. In case of people having eye defects, the spectacles form the virtual image of the object and the eye lens form the real and inverted image on the retina.

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Question 552 Marks

Can a virtual image be photographed by a camera?

Answer

Yes, when you stand in front of plane mirror image is virtual and can be photographed.

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Question 562 Marks

An optical fibre $(\mu=1.72)$ is surrounded by a glass coating $(\mu=1.50).$ Find the critical angle for total internal/ reflection at the fibre-glass interface.

Answer

For calculation of critical angle,
$\frac{\sin\text{i}}{\sin\text{r}}=\frac{\mu_2}{\mu_1}$
$\Rightarrow\frac{\sin\theta_\text{C}}{\sin90^\circ}=\frac{15}{1.72}$
$=\frac{75}{86}$
$\Rightarrow\theta_\text{C}=\sin^{-1}\Big(\frac{75}{86}\Big)$

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Question 572 Marks

A nearsighted person cannot clearly see beyond 200cm. Find the power of the lens needed to see objects at large distances.

Answer

For the near sighted person,$\text{u}= \infty$ and $\text{v}=-200\text{cm}=-2\text{m}$
So, $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{-2}-\frac{1}{\infty}$
$\frac{1}{\text{f}}=-\frac{1}{2}=-0.5$
So, power of the lens is -0.5D

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Question 582 Marks

A convex lens made of a material of refractive index $n_1$ is kept in a medium of refractive index $n_2$. Parallel rays of light are incident on the lens. Complete the path of rays of light emerging from the convex lens if: (i) $\mathrm{n}_1>\mathrm{n}_2$ (ii) $\mathrm{n}_1=\mathrm{n}_2$ (iii) $\mathrm{n}_1<\mathrm{n}_2$.
$\frac{1}{\text{f}}=\Big(\frac{\text{n}_1}{\text{n}_2}-1\Big)\Big(\frac{1}{\text{R}_2}+\frac{1}{\text{R}_2}\Big)$

Answer

In case (i) $\mathrm{n}_1>\mathrm{n}_2$, the lens behaves as convergent lens.
In case (ii) $n_1=n_2$, the lens behaves as a plane plate.
In case (iii) $\mathrm{n}_1<\mathrm{n}_2$, the lens behaves as a divergent lens.

The path of rays in all the three cases is shown in fig.

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Question 592 Marks

A point source is placed at a depth h below the surface of water (refractive index = μ).

Show that light escapes through a circular area on the water surface with its centre directly above the point source.
Find the angle subtended by a radius of the area on the source.

Answer

Let, x = radius of the circular area

$\frac{\text{x}}{\text{h}}=\tan\theta_\text{C}$ (where C is the critical angle)

$\Rightarrow\frac{\text{x}}{\text{h}}=\frac{\sin\theta_\text{C}}{\sqrt{1-\sin^2\theta_\text{C}}}=\frac{\frac{1}{\mu}}{\sqrt{1-\frac{1}{\mu^2}}} \ \Big(\because \ \sin\theta_\text{C}=\frac{1}{\mu}\Big)$

$\Rightarrow\frac{\text{x}}{\text{h}}=\frac{1}{\sqrt{\mu^2-1}}$ or $\text{x}=\frac{\text{h}}{\sqrt{\mu^2-1}}$

So, light escapes through a circular area on the water surface directly above the point source.

Angle subtained by a radius of the area on the source, $\theta_\text{C}=\sin^{-1}\Big(\frac{1}{\mu}\Big)$

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Question 602 Marks

Light falls from glass $(\mu=1.5)$ to air. Find the angle of incidence for which the angle of deviation is 90°.

Answer

Since, $\mu=1.5,$ Critial angle $=\sin^{-1}\Big(\frac{1}{\mu}\Big)=\sin^{-1}\Big(\frac{1}{1.5}\Big)=41.8^{\circ}$
We know, the maximum attainable deviation in refraction is (90° - 41.8°) = 47.2°
So, in this case, total internal reflection must have taken place.
In reflection,
Deviation = 180° - 2i = 90°
⇒ 2i = 90° ⇒ i = 45°.

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Question 612 Marks

A simple microscope is rated 5 for a normal relaxed eye. What will be its magnifying power for a relaxed farsighted eye whose near point is 40cm?

Answer

The simple microscope has magnification of 5 for normal relaxed eye (D = 25cm).Because, the eye is relaxed the image is formed at infinity (normal adjustment)
So, $\text{m}=5=\frac{\text{D}}{\text{f}}=\frac{25}{\text{f}}\Rightarrow\text{f}=5\text{cm}$
For the relaxed farsighted eye, $\text{D} = 40\text{cm}$
So, $\text{m}=\frac{\text{D}}{\text{f}}=\frac{40}{5}=8$
So, its magnifying power is 8X.

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Question 622 Marks

A concave mirror and a converging lens have the same focal length in air. Which one of the two will have greater focal length when both are immersed in water?

Answer

Converging lens; the focal length of a spherical mirror remains unaffected.
For converging lens $\frac{1}{\text{f}}=\Big(\frac{\text{n}_1}{\text{n}_2}-1\Big)\Big(\frac{1}{\text{R}_2}-\frac{1}{\text{R}_2}\Big)$
When it is immersed in water $\mathrm{n}_2$ (in water) $>\mathrm{n}_2$ (air)
$\Big(\frac{\text{n}_1}{\text{n}_2}-1\Big)$ decreases hence focal length of converging lens increases in water.

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Question 632 Marks

A single lens is mounted in a tube. A parallel beam enters the tube and emerges out of the tube as a divergent beam. Can you say with certainty that there is a diverging lens in the tube?

Answer

No,

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Question 642 Marks

A farsighted person cannot see objects placed closer to 50cm. Find the power of the lens needed to see the objects at 20cm.

Answer

For the far sighted person,u = -20cm, v = -50cm
from lens formula $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{f}}=\frac{1}{-50}-\frac{1}{-20}=\frac{1}{-20}-\frac{1}{50}=\frac{3}{100}$
$\Rightarrow\text{f}=\frac{100}{3}\text{cm}=\frac{1}{3}\text{m}$
So, power of the lens $=\frac{1}{\text{f}}=3\text{ Diopter}$

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Question 652 Marks

A convex lens (n = 1.5) of focal length fa is immersed.

In water n = 1.33 and.
In carbon disulphide n = 1.6, how does the lens behave in the two cases?

Answer

When lens is immersed in water, it behaves as a convex lens but its focal length will increase.
When convex lens is immersed in carbon-disulphide, it will behave as a concave lens.

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Question 662 Marks

Can you ever have a situation in which a light ray goes undeviated through a prism?

Answer

Yes.

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Question 672 Marks

The equation $\omega=\frac{\mu_\text{v}-\mu_\text{r}}{\mu-1}$ asderived for a prism having small refracting angle. Is it also valid for a prism of large refracting angle? Is it also valid for a glass slab or a glass sphere?

Answer

Dispersive power depends on angular deviation, and angular deviation is valid only for a small refracting angle and a small angle of incidence. Therefore, dispersive power is not valid for a prism of large refracting angle. It is also not valid for a glass slab or a glass sphere, as it has a large refracting angle.

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Question 682 Marks

A compound microscope forms an inverted image of an object. In which of the following cases it is kely to create difficulties?

Looking at small germs.
Looking at circular spots.
Looking at a vertical tube containing some water.

Answer

Looking at a vertical tube containing some water

Explantion:

If the experimentalist is looking at a vertical tube containing some water, he has to be careful, as the lower meniscus will appear as upper.

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Question 692 Marks

A spherical surface of radius 30cm separates two transparent media A and B with refractive indices 1.33 and 1.48 respectively. The medium A is on the convex side of the surface. Where should a point object be placed in medium A so that the paraxial rays become parallel after refraction at the surface?

Answer

Since, paraxial rays become parallel after refraction i.e. image is formed at $\infty.$
$\text{v}=\infty, \ \mu_1=1.33, \ \text{u}=?, \ \mu_2=1.48, \ \text{R}=30\text{cm}$
$\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$
$\Rightarrow\frac{1.48}{\infty}-\frac{1.33}{\text{u}}=\frac{1.48-1.33}{30}\Rightarrow-\frac{1.33}{\text{u}}-\frac{0.15}{30}$
$\Rightarrow\text{u}=-266.0\text{cm}$
$\therefore$ Object should be placed at a distance of 266cm from surface (convex) on side A.

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Question 702 Marks

The refractive index of a material changes by 0.014 as the colour of the light changes from red to violet. A rectangular slab of height 2.00cm made of this material is placed on a newspaper. When viewed normally in yellow light, the letters appear 1.32cm below the top surface of the slab. Calculate the dispersive power of the material.

Answer

Given that, $\mu_\text{v}-\mu_\text{r}=0.014$
$\text{Again,}\ \mu_\text{y}=\frac{\text{Real depth}}{\text{Apparent depth}}=\frac{2.00}{1.30}=1.515$
So, dispersive power $=\frac{\mu_\text{v}-\mu_\text{r}}{\mu_\text{y}-1}=\frac{0.014}{1.515-1}=0.027$

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Question 712 Marks

Can the dispersive power $\omega=\frac{\mu_\text{v}-\mu\text{r}}{\mu-1}$ be negative? What is the sign of co if a hollow prism is immersed into water?

Answer

No, it cannot be negative, as the refractive index for violet light is always greater than that for red light. Also, refractive index is inversely proportional to $\lambda^2$. The sign of wwill be positive, as $\mu$ is still greater than 1 and es $\mu_\text{v}>\mu_\text{r}.$

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Question 722 Marks

The minimum deviations suffered by red, yellow and violet beams passing through an equilateral transparent prism are 38.4°, 38.7° and 39.2° respectively. Calculate the dispersive power of the medium.

Answer

Given that, $\delta_\text{r}=38.4^\circ,\delta_\text{y}=38.7^\circ\ \text{and}\ \delta_\text{v}=39.2^\circ$
Dispersive power $=\frac{\mu_\text{v}-\mu_\text{r}}{\mu_\text{y}-1}=\frac{(\mu_\text{v}-1)-(\mu_\text{r}-1)}{(\mu_\text{y}-1)}$ $=\frac{\Big(\frac{\delta_\text{v}}{\text{A}}\Big)-\Big(\frac{\delta_\text{r}}{\text{A}}\Big)}{\Big(\frac{\delta_\text{v}}{\text{A}}\Big)}\ [\because\delta=(\mu-1)\text{A}]$
$=\frac{\delta_\text{v}-\delta_\text{r}}{\delta_\text{y}}=\frac{39.2-3.84^\circ}{38.7}=0.0204$

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Question 732 Marks

A simple microscope using a single lens often shows coloured image of a white source. Why?

Answer

A simple microscope consists of a single convex lens. Sometimes due to chromatic and spherical aberrations, the image of a white source seems coloured at the corners of the lens and somewhere in between.

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Question 742 Marks

The refractive index of a material of a concave lens is $n_1$. It is immersed in a medium of refractive index $n_2 A$ parallel beam of light is incident on the lens. Trace the path of emergent rays when (i) $n_2=n_1$ (ii) $n_2>n_1$ (iii) $n_2<n_1$.
$\frac{1}{\text{f}}=\Big(\frac{\text{n}_1}{\text{n}_2}-1\Big)\Big(\frac{1}{\text{R}_2}-\frac{1}{\text{R}_2}\Big)$

Answer

$\text{For} \text{ n}_1 = \text{n}_2\text{ f} = \infty$
$\text{For} \text{ n}_1 < \text{n}_2 \text{ f} > 0$
$\text{For} \text{ n}_1 > \text{n}_2 \text{ f} < 0$
The path of rays in three cases is shown in fig.

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Question 752 Marks

Suggest a method to produce a rainbow in your house.

Answer

A rainbow can be produced using a prism. Another way of producing a rainbow is to dip a mirror inside water, keeping it inclined along the wall of a tumbler. The light coming from water after reflecting from the mirror will give a rainbow.

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Question 762 Marks

The angular magnification of a system is less than one. Does it mean that the image formed is inverted?

Answer

No, angular magnification is the ratio of the angle subtended by the final image on the eye to the angle subtended by the object on the unaided eye. Its value less than one signifies reduction in the size of the image. It does not mean that the image is inverted.

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Question 772 Marks

Find the maximum angle of refraction when a light ray is refracted from glass $(\mu=1.50)$ to air.

Answer

From the definition of critical angle, if refracted angle is more than 90°, then reflection occurs, which is known as total internal reflection.
So, maximum angle of refraction is 90°.

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Question 782 Marks

If an object far away from a convex mirror moves towards the mirror, the image also moves. Does it move faster, slower or at the same speed as compared to the object?

Answer

Slower $\text{V}_{\text{img}}=\text{m}^2\text{v}_0$ [m < 1 as far away]

It objects moves from infinity to 2f as distance moved by object is more in same time hence velocity of object is more.

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Question 792 Marks

A Galilean telescope is 27cm long when focused to form an image at infinity. If the objective has a focal length of 30cm, what is the focal length of the eyepiece?

Answer

For the given Galilean telescope, (When the image is formed at infinity) $\mathrm{f}_0=30 \mathrm{~cm}, \mathrm{~L}=27 \mathrm{~cm}$ Since $L=f_0-\left|f_e\right|$
[Since, concave eyepiece lens is used in Galilean Telescope]
$\Rightarrow f_e=f_0-L=30-27=3 \mathrm{~cm}$

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Question 802 Marks

A light ray is incident normally on the face AB of a right-angled prism ABC $(\mu=1.50)$ as shown in figure. What is the largest angle $\phi$ for which the light ray is totally reflected at the surface AC?

Answer

Let $\theta_\text{c}$ be the critical angle for the glass
$\frac{\sin\theta_{\text{c}}}{\sin90^{\circ}}=\frac{1}{\text{x}}\Rightarrow\sin\theta_{\text{c}}=\frac{1}{1.5}=\frac{2}{3}\Rightarrow\theta_{\text{c}}=\sin^{-1}\Big(\frac{2}{3}\Big)$
From figure, for total internal reflection, $90^{\circ}-\phi>\theta_{\text{c}}$
$\Rightarrow\phi<90^{\circ}-\theta_{\text{c}}\Rightarrow\phi<\cos^{-1}\Big(\frac{2}{3}\Big)$
So, the largest angle for which light is totally reflected at the surface is $\cos^{-1}\Big(\frac{2}{3}\Big).$

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Question 812 Marks

An extended object is placed at a distance of 5.0cm from a convex lens of focal length 8.0cm.

Draw the ray diagram (to the scale) to locate the image and from this, measure the distance of the image from the lens.
Find the position of the image from the lens formula and see how close the drawing is to the correct result.

Answer

Given, u = -5cm, f = 8cm

So, $\Rightarrow\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{8}-\frac{1}{5}=\frac{-3}{40}$
$\Rightarrow\text{v}=-13.3\text{cm}$ (virtual image).

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Question 822 Marks

A telescope has been adjusted for relaxed eye. You are asked to adjust it for least distance of distinct vision, then how will you change the distance between two lenses?

Answer

For relaxed eye, $\text{L} = \text{f}_0 +\text{f}_{\text{e}}$
For least distance of distinct vision $\text{L}' = \text{f}_0 + \text{u}_{\text{e}}, \text{u}_{\text{e}} < \text{f}_{\text{e}}$
Therefore, L' < L, that is, the distance will be decreased.

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Question 832 Marks

Why is there no dispersion in the light refracted through a rectangular glass slab?

Answer

It can be considered to be equivalent to two prisms in reverse way. In the case of a rectangular glass slab the rays of all colours combine together in the reverse prism. Hence, there is no dispersion.

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Question 842 Marks

An object is placed at the principal focus of a concave lens of focal length f. Where will its image be formed?
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{f}}+\frac{1}{\text{u}}$

Answer

Here u = -f and for a concave lens f = -f
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{f}}+\frac{1}{\text{u}}$
That image will be formed between optical centre and focus of lens; towards the side of the object.

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Question 852 Marks

A person wears glasses of power -2.5D. Is the person farsighted or nearsighted? What is the far point of the person without the glasses?

Answer

The person wears glasses of power -2.5DSo, the person must be near sighted
$\text{u}=\infty,$ $\text{v}=\text{far point,}$ $\text{f}=\frac{1}{-2.5}=-0.4\text{m}=-40\text{cm}$
Now, $\frac{1}{\text{v}}-\frac{1}{\text{v}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{u}}+\frac{1}{\text{f}}=0+\frac{1}{-40}$
$\Rightarrow \text{v}=-40\text{cm}$
So, the far point of the person is 40cm.

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Question 862 Marks

The focal length of a convex lens made of glass is 20cm. What will be its new focal length when placed in a medium of refractive index 1.25?

Answer

$\frac{1}{\text{f}}=(\mu-1)\Big[\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big]$
$\Rightarrow\frac{1}{20}=\frac{1}{4}\Big[\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big]$
$\frac{1}{\text{f}}=\frac{1}{4}\Big[\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big]$
$\Rightarrow\text{f}'=40\text{cm}$

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Question 872 Marks

The equation of refraction at a spherical surface is $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}-=\frac{\mu_2-\mu_1}{\text{R}}.$ Taking $\text{R}=\infty,$ show that this equation leads to the equation $\frac{\text{Real depth}}{\text{Apparent depth}}=\frac{\mu_2}{\mu_1}$ for refraction at a plane surface.

Answer

$\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}-=\frac{\mu_2-\mu_1}{\infty}$
$\frac{\mu_2}{\text{v}}=\frac{\mu_1}{\text{u}}$
$\Rightarrow\frac{\mu_2}{\mu_1}=\frac{\text{v}}{\text{u}}.$

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Question 882 Marks

Light is incident from glass $(\mu=1.5)$ to air. Sketch the variation of the angle of deviation $\delta$ with the angle of incident i for 0 < i < 90°.

Answer

Refractive index of glass $\mu_{\text{g}}=1.5$
Given, 0° < i < 90°
Let, $\theta_\text{C}$ → Critical angle.
$\frac{\sin\theta_\text{C}}{\sin\text{r}}=\frac{\mu_{\text{a}}}{\mu_{\text{g}}}$
$\Rightarrow\frac{\sin\theta_\text{C}}{\sin90^{\circ}}=\frac{1}{1.5}=0.66$
$\Rightarrow\text{C}=40^{\circ}48''$
The angle of deviation due to refraction from glass to air increases as the angle of incidence increases from 0° to 40°48''. The angle of deviation due to total internal reflection further increases for 40°48'' to 45° and then it decreases.

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Question 892 Marks

A right-angled crown glass prism with critical angle 41° is placed before an object, PQ in two positions as shown in the figures (i) and (ii). Trace the paths of the rays from P and Q passing through the prisms in the two cases.

Answer

The formation of images is shown in figures (i) and (ii).

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Question 902 Marks

The eyepiece of an astronomical telescope has a focal length of 10cm. The telescope is focused for normal vision of distant objects when the tube length is 1.0m. Find the focal length of the objective and the magnifying power of the telescope.

Answer

For the given astronomical telescope in normal adjustment, $\mathrm{F}_{\mathrm{e}}=10 \mathrm{~cm}, \mathrm{~L}=1 \mathrm{~m}=100 \mathrm{~cm}$
So $, \mathrm{f}_0=\mathrm{L}-\mathrm{f}_{\mathrm{e}}=100-10=90 \mathrm{~cm}$
and, magnifying power $=\frac{\text{f}_0}{\text{f}_\text{e}}=\frac{90}{10}=9$

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Question 912 Marks

If three identical prisms are combined, is it possible to pass a beam that emerges undeviated? Undispersed?

Answer

No, it is not possible even when prisms are be combined with their refractive angle reversed with respect to each other. There will be at least a net deviation and dispersion equal to the dispersion and deviation produced by a single prism.

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Question 922 Marks

"Monochromatic light should be used to produce pure spectrum". Comment on this statement.

Answer

No, monochromatic light cannot be used to produce a pure spectrum. A spectrum is produced when a light of different wavelengths is deviated through different angles and gets separated. Monochromatic light, on the other hand, has a single wavelength.

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Question 932 Marks

How does the power of a convex lens vary, if the incident red light is replaced by violet light?

Answer

Power of a lens increases if red light is replaced by violet light because.
$\text{P}=\frac{1}{\text{f}}=(_{\text{a}\text{n}_{\text{g}}-1})\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_1}\Big)$ and refractive index is maximum for violet light in visible region of spectrum.

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