M.C.Q [1M]

Question 11 Mark

Two concave lenses L₁ and L₂ are kept in contact with each other. If the space between the two lenses is filled with a material of refractive index $\mu\approx1,$ the magnitude of the focal length of the combination:

Becomes undefined.
Remains unchanged.
Increases.
Decreases.

Answer

Decreases.

Explanation:

$\frac{1}{\text{f}}=\frac{1}{\text{f}_{\text{L}_1}}+\frac{1}{\text{f}_{\text{L}_2}}$

$\frac{1}{\text{f}_{\text{L}_1}}=(\mu-1)\Big(\frac{-2}{\text{R}}\Big)=\frac{1}{\text{f}_{\text{L}_2}}$

Local length of the combination.

$\frac{1}{\text{f}}=(\mu-1)\Big(\frac{-2}{\text{R}}\Big)+(\mu-1)\Big(\frac{-2}{\text{R}}\Big)$

$\frac{1}{\text{f}}=-4(\mu-1)\Big(\frac{1}{\text{R}}\Big)$

$\text{f}=\frac{\text{R}}{4(\mu-1)}$

Where $\text{f}_{\text{L}_1}=\text{f}_{\text{L}_2}=\frac{\text{R}}{2(\mu-1)}$

$(\text{f}_{\text{L}_1}=\text{f}_{\text{L}_2})>\text{f}$

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Question 21 Mark

Two rays A and B being reflected by a mirror and going as A' and B'. The mirror:

Is plane.
Is convex.
Is concave.
May be any spherical mirror.

Answer

Is plane.

Explanation:

Here initially A & B is parallel to each other after reflection by teh plane mirror A' & B' goes Parallel to each other.

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Question 31 Mark

Three transparent media of refractive indices $\mu_1,\mu_2$ and $\mu_3.$ A point object O is placed in the medium $\mu_2.$ If the entire medium on the right of the spherical surface has refractive index $\mu_1,$ the image forms at O'. If this entire medium has refractive index $\mu_3,$ the image forms at O". In the situation shown:

The image forms between O' and O"
The image forms to the left of O'
The image forms to the right of O"
Two images form, one at O' and the other at O".

Answer

Two images form, one at O' and the other at O".

Explanation:

m₁, Image is O¹

m₃, Image is O¹¹

Spherical Surface formula

$\frac{\mu^{11}}{\text{v}}-\frac{\mu_1}{\mu}=\frac{\mu^{11}-\mu^1}{\text{R}}$

If ray goes to m² to m¹ than Image is formed at O¹ and if ray goes to m₂ to m₃ than Image is formed at O¹¹.

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Question 41 Mark

A point object is placed at a distance of 30cm from a convex mirror of focal length 30cm. The image will form at:

Infinity.
Pole.
Focus.
15cm behind the mirror.

Answer

Infinity.

Explanation:

By mirror formula:

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

Here u = -30cm

f = +30cm

So, $\frac{1}{\text{V}}-\frac{1}{30}=\frac{1}{30}$

Þ v= 15cm behind the mirror.

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Question 51 Mark

A point object O is placed on the principal axis of a convex lens of focal length f = 20cm at a distance of 40cm to the left of it. The diameter of the lens is 10cm. An eye is placed 60cm to right of the lens and a distance h below the principal axis. The maximum value of h to see the image is:

0
2.5cm
5cm
10cm.

Answer

2.5cm

Explanation:

$\tan\theta=\frac{5}{40}=\frac{\text{h}}{20}$

$\text{h}=\frac{5}{2}=\frac{\text{h}}{20}$

$\text{h}=\frac{5}{2}=2.5\text{cm}$

The maximum value of "h =2.5cm" to see the Image of the object.

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Question 61 Mark

Total internal reflection can take place only if:

Light goes from optically rarer medium (smaller refractive index) to optically denser medium.
Light goes from optically denser medium to rarer medium.
The refractive indices of the two media are close to each other.
The refractive indices of the two media are widely different.

Answer

Light goes from optically denser medium to rarer medium.

Explanation:

T.I.R. (Total Internal reflection)

i < C (condition for T.I.R.)

By Snell's Law

$\text{n}_1\sin\text{i}=\text{n}_2\sin90^{\circ}$

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

$\text{þ} \ \text{C}=\sin^{-1}$

$\because \ -1\leq\sin\text{i}\leq1$

n₁ > n₂ (so we can conclude that light goes from optically denser medium to rarer medium & incident angle is greater than the critical angle.)

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Question 71 Mark

The image of an extended object, placed perpendicular to the principal axis of a mirror, will be erect if:

The object and the image are both real.
The object and the image are both virtual.
The object is real but the image is virtual.
The object is virtual but the image is real.

Answer

The object is real but the image is virtual.
The object is virtual but the image is real.

Explanation:

$\text{m}=\frac{-\text{v}}{\mu}=\frac{\text{h}_1}{\text{h}_0}$ (for mirror)

Image will be erect mean height of the object and Image will e lies in same side. It mean if object isreal then Image in virtual. If object is virtula then Image is real.

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Question 81 Mark

A convex lens forms a real image of a point object placed on its principal axis. If the upper half of the lens is painted black:

The image will be shifted downward.
The image will be shifted upward.
The image will not be shifted.
The intensity of the image will decrease.

Answer

The image will not be shifted.
The intensity of the image will decrease.

Explanation:

By lens formula

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

Due to Black Pointed focal Lenght of the Lens will not change.

So Image will not be shifted due to Black point. But Intensily of Image will decrease.

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Question 91 Mark

Consider three converging lenses L₁, L₂ and L₃ having identical geometrical construction. The index of refraction of L₁ and L₂ are $\mu_1$ and $\mu_2$ respectively. The upper half of the lens L₃ has a refractive index $\mu_1$ and the lower half has $\mu_2$. A point object O is imaged at O₁ by the lens L₁ and at O₂ by the lens L₂ placed in same position. If L₃ is placed at the same place:

There will be an image at O₁
There will be an image at O₂
The only image will form somewhere between O₁ and O₂
The only image will form away from O_2.

Answer

There will be an image at O₁
There will be an image at O₂

Explanation:

It rays are Passing through m₁ then Image will be form at "O₁" and If rays are Passing through m₂ then Image will be form at "O₂".

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Question 101 Mark

A convex lens is made of a material having refractive index 1.2. Both the surfaces of the lens are convex. If it is dipped into water $(\mu=1.33),$ it will behave like:

A convergent lens.
A divergent lens.
A rectangular slab.
A prism.

Answer

a divergent lens.

Explanation:

Here P, P₁ & P₂ are the Power of Lenses.

P = P₁ + P₂

$\frac{1}{\text{f}}=\frac{1}{\text{f}_1}=\frac{1}{\text{f}_2}$

$(\mu-1)\Big(\frac{2}{\text{R}}\Big)+(\mu'-1)\Big(\frac{-1}{\text{R}}\Big)$

$(1.2-1)\Big(\frac{2}{\text{R}}\Big)-\Big(\frac{4}{3}-1\Big)\Big(\frac{1}{\text{R}}\Big)$

$\frac{1}{\text{f}}=\frac{2}{5\text{R}}-\frac{1}{3\text{R}}$

$\frac{1}{\text{f}}=\frac{6-5}{15\text{R}}$

$\text{f}=15\text{R}$

Focal lenght of combined is positive, but it's magnitude in capair to f₁ & f₂ is High. So it will be hare like a divergent lens.

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Question 111 Mark

A point source of light is placed in front of a plane mirror:

All the reflected rays meet at a point when produced backward.
Only the reflected rays close to the normal meet at a point when produced backward.
Only the reflected rays making a small angle with the mirror, meet at a point when produced backward.
Light of different colours make different images.

Answer

All the reflected rays meet at a point when produced backward.

Explanation:

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Question 121 Mark

A screen is placed a distance 40cm away from an illuminated object. A converging lens is placed between the source and the screen and it is attempted to form the image of the source on the screen. If no position could be found, the focal length of the lens:

Must be less than 10cm.
Must be greater than 20cm.
Must not be greater than 20cm.
Must not be less than 10cm.

Answer

Must be greater than 20cm.

Explanation:

$\text{v}=(40-4)$

$\frac{1}{\text{f}}=\frac{1}{40-4}-\frac{1}{(-\text{u})}$

$\frac{\text{df}}{\text{du}}=0$ for f minimum.

$\frac{\text{df}}{\text{du}}=1-\frac{\text{u}}{20}=0$

$\text{u}=20$

$\text{f}_{\text{min}}=10\text{cm}$

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Question 131 Mark

If the light moving in a straight line bends by a small but fixed angle, it may be a case of:

Reflection.
Refraction.
Diffraction.
Dispersion.

Answer

Reflection.
Refraction.

Explanation:

When the light strikes on a surface nearly parallel to it, it then bends by a small and fixed angle after reflection. Also, when the light travels from one medium to another with slight differences in their refractive indices, it bends by a small angle. Thus, the bending of light by a small but fixed angle can be the case of either reflection or refraction.

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Question 141 Mark

Mark the correct options:

If the incident rays are converging, we have a real object.
If the final rays are converging, we have a real image.
The image of a virtual object is called a virtual image.
If the image is virtual, the corresponding object is called a virtual object.

Answer

If the final rays are converging, we have a real image.

Explanation:

This is because a real image is formed by converging reflected/ refracted rays from a mirror/ lens.

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Question 151 Mark

Two rays A and B being reflected by a mirror and going as A' and B'. The mirror:

Is plane.
Is convex.
Is concave.
May be any spherical mirror.

Answer

Is plane.

Explanation:

Here initially A & B is parallel to each other after reflection by teh plane mirror A' & B' goes Parallel to each other.

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Question 161 Mark

The rays of different colours fail to converge at a point after going through a converging lens. This defect is called:

Spherical aberration.
Distortion.
Coma.
Chromatic aberration.

Answer

Chromatic aberration.

Explanation:

The rays of different colours fail to converge at a Point after going through a converging Lens. This defect is called chromatic oberration.

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Question 171 Mark

Which of the following (referred to a spherical mirror) do (does) not depend on whether the rays are paraxial or not?

Pole.
Focus.
Radius of curvature.
Principal axis.

Answer

Pole.
Focus.

Principal axis.

Explanation:

If Paraxial rays comes to parallel to the spherical mirror is pasees to the Focus of the spherical mirror.

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Question 181 Mark

In image formation from spherical mirrors, only paraxial rays are considered because they:

Are easy to handle geometrically.
Contain most of the intensity of the incident light.
Form nearly a point image of a point source.
Show minimum dispersion effect.

Answer

Form nearly a point image of a point source.

Explanation:

In Image formation from spherical mirrors, only paraxial rays are considered because they form nearly a point Image of a point source. Angle of Incidence of Paraxial rays is very small.

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Question 191 Mark

The image formed by a concave mirror:

Is always real.
Is always virtual.
Is certainly real if the object is virtual.
Is certainly virtual if the object is real.

Answer

Is certainly real if the object is virtual.

Explanation:

Object	Image
O₁	I₁
O₂	I₂
O3	I₃
O₄	I₄

Here O₄ is virtual oblect & I₄ is real image.

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Question 201 Mark

A symmetric double convex lens is cut in two equal parts by a plane perpendicular to the principal axis. If the power of the original lens was 4D, the power of a cutlens will be:

Answer

Explanation:

Before cut

$\frac{1}{\text{f}}=(\mu-1)\Big(\frac{2}{\text{R}}\Big)=4\text{D} \ ...(1)$

After cut

$\frac{1}{\text{f}_1}=(\mu-1)\Big(\frac{1}{\text{R}}\Big)+\frac{1}{\text{f}_2}=(\mu-1)\Big(\frac{1}{2}\Big) \ ...(2)$

From eq. (1) we get Power of f₁ = power of f₂

$\text{P}=\frac{1}{\text{f}_1}=\frac{1}{\text{f}_2}=2\text{D}$

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Question 211 Mark

A parallel beam of light is incident on a converging lens parallel to its principal axis. As one moves away from the lens on the other side on its principal axis, the intensity of light:

Remains, constant.
Continuously increases.
Continuously decreases.
First increases then decreases.

Answer

First increases then decreases.

Explanation:

The intensity o light is first increases then decreases.

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Question 221 Mark

A symmetric double convex lens is cut in two equal parts by a plane containing the principal axis. If the power of the original lens was 4D, the power of a divided lens will be:

Answer

Explanation:

Before cut

$\frac{1}{\text{f}}=(\mu-1)\Big(\frac{2}{\text{R}}\Big)=4\text{D}$

After cut

$\frac{1}{\text{f}_1}=(\mu-1)\Big(\frac{1}{\text{R}}\Big)=\text{P}_1$

$\&\frac{1}{\text{f}_2}=(\mu-1)\Big(\frac{1}{\text{R}}\Big)=\text{P}_2$

Power of a divided lens will be = P₁ + P₂

$=(\mu-1)\Big(\frac{2}{\text{R}}\Big)$

$=4\text{D}$

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Question 231 Mark

A thin lens is made with a material having refractive index $\mu=1.5.$ Both the sides are convex. It is dipped in water $(\mu=1.33).$ It will behave like:

A convergent lens.
A divergent lens.
A rectangular slab.
A prism.

Answer

a convergent lens.

Explanation:

Here P, P₁ & P₂ are the power of Lenses.

P = P₁ + P₂

$\frac{1}{\text{f}}=\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}$

$=(\mu-1)\Big(\frac{2}{\text{R}}\Big)+(\mu'-1)\Big(\frac{-1}{\text{R}}\Big)$

$=\Big(\frac{3}{2}-1\Big)\Big(\frac{2}{\text{R}}\Big)-\Big(\frac{4}{3}-1\Big)\Big(\frac{1}{\text{R}}\Big)$

$\frac{1}{\text{f}}=\frac{1}{\text{R}}-\frac{1}{3\text{R}}$

$\frac{1}{\text{f}}=\frac{3-1}{3\text{R}}$

$\text{f}=\frac{3\text{R}}{2}$

focal length of combined is positive means it will behave like a canvergent lens.

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Question 241 Mark

Four modifications are suggested in the lens formula to include the effect of the thickness t of the lens. Which one is likely to be correct?

$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{uf}}$
$\frac{1}{\text{v}^2}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}-\text{t}}-\frac{1}{\text{u}+\text{t}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}-\frac{1}{\text{u}}+\frac{\text{t}}{\text{uv}}=\frac{\text{t}}{\text{f}}$

Answer

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

Explanation:

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Question 251 Mark

A point source of light is placed at a distance, of 2f from a converging lens of focal length f. The intensity on the other side of the lens is maximum at a distance: