Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 62

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Question 1
1 / -0

The magnification of the image when an object is placed at a distance x from the principle focus of a mirror of focal length f is?

A
$$\displaystyle\frac{x}{f}$$

B
$$1+f/x$$

C
$$\displaystyle\frac{f}{x}$$

D
$$1-\displaystyle\frac{f}{x}$$

Solution

$$\textbf{Step 1: Sign Convention and position of Object [Refer fig.]} $$
Pole of the mirror is taken as origin, and all distances are taken from pole.

Sign convention $$\rightarrow $$ Direction of incident ray is +ve, i.e. rays coming from object to mirror, rightwards here.

As Object$$(O)$$ is leftwards, So object distance $$ u= -(x + f) $$
As focus is also to the left of the pole, So focal length $$=-f$$

$$\textbf{Step 2: From mirror formula} $$

$$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} $$

$$\dfrac{1}{-f} = \dfrac{1}{v} + \dfrac{1}{-(x + f)} $$

$$\dfrac{1}{v} = \dfrac{1}{x + f} - \dfrac{1}{f} $$

$$\dfrac{1}{v} = \dfrac{f - x - f}{f (x + f)} $$

$$v = - \dfrac{f(x + f)}{x} $$

$$\textbf{Step 2: Magnification of mirror} $$

$$m = -\dfrac{v}{u} $$

$$m = \dfrac{\dfrac{-f (x + f)}{x}}{- (x + f)} $$ $$=\ \ -\dfrac{f}{x} $$

$$\therefore\ \ \ |m| = \dfrac{f}{x}$$

Hence, Option $$C$$ is correct
Question 2
1 / -0

A real object is placed at a distance $$f$$ from the pole of a convex mirror, in front of the convex minor. If focal length of the mirror is $$f$$, then distance of the image from the pole of the mirror is :

A
$$2\, f$$

B
$$\frac{f}{2}$$

C
$$4\, f$$

D
$$\frac{f}{4}$$

Solution

Here the object distance is $$f$$,
So $$u=-f$$ and $$F=f$$
Using the mirror's formula $$\dfrac{1}{F}=\dfrac{1}{v}+\dfrac{1}{u}$$
$$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{-f}$$
$$\implies \ v=\dfrac{f}{2}$$
Question 3
1 / -0

The refractive index of glass with respect to air is $$\dfrac{3}{2}$$ and that of water with respect to air is $$\dfrac{3}{2}$$. What is the refractive index of glass with respect to water?

A
$$9 : 8$$

B
$$1 : 2$$

C
$$2 : 1$$

D
$$8 : 9$$

Solution

Given,

Refractive index of glass=$$\dfrac{3}{2}$$

Refractive index of water=$$\dfrac{4}{3}$$

Refractive Index of glass with respect to water= $$\dfrac{3/2}{4/3}$$
=$$\dfrac{9}{8}$$
Question 4
1 / -0

You are given two lenses, a converging lens with focal length + 10 cm and a diverging lens with focal length -20 cm. Which of the following would produce a real image that is smaller than the object?

A
Placing the object 5 cm from the converging lens

B
Placing the object 15 cm from the converging lens

C
Placing the object 25 cm from the converging lens

D
Placing the object 15 cm from the diverging lens

Solution

When the object is placed beyond the center of curvature of a converging lens, a real and diminished image is formed between the focus and the center of curvature.
The focal length of the converging lens is $$+10\ cm$$
The radius of curvature of the converging lens is $$2f=+20\ cm$$
Thus, the object must be placed beyond $$20\ cm$$ or at $$25\ cm$$.
Question 5
1 / -0

Which among these is the angle of incidence?

A
A

B
B

C
C

D
D

Solution

The angle of incidence is the angle between incident ray and the normal. So, the right answer is B.
Question 6
1 / -0

Magnification produced by a convex mirror is $$\dfrac{1}{3}$$, then distance of the object from mirror is

A
$$\dfrac{f}{3}$$

B
$$\dfrac{2f}{3}$$

C
$$1f$$

D
$$2f$$

Solution

$$ \dfrac{1}{f} = \dfrac{1}{v}-\dfrac{1}{u} $$, multiply using by u
$$ \Rightarrow \dfrac{u}{f} = \dfrac{u}{v}-\dfrac{u}{u} = \dfrac{u}{v}-1 $$
$$ \Rightarrow \dfrac{u}{f}+1 = \dfrac{u}{v}\Rightarrow \dfrac{u}{v} = \dfrac{u+f}{f} $$
$$ \Rightarrow \dfrac{v}{u} = \dfrac{f}{f+u}\Rightarrow m|\dfrac{-v}{u}| = \dfrac{f}{f+u} $$
given that $$ m = \dfrac{1}{3} $$
So $$ \dfrac{1}{3} = \dfrac{f}{f+u}\Rightarrow f+u = 3f $$
$$ \Rightarrow \boxed{u = 2f} $$
Question 7
1 / -0

The phenomenon of bending of light at the surface of separation of two media is called :

A
Refraction of light

B
Reflection of light

C
Deflection of light

D
Absorption of light

Solution

When light travels from one medium to another, the light gets bent and this phenomenon is called refraction of light. The bending of light is due to the different speeds of light in different media.
Question 8
1 / -0

A ray of light strikes a glass slab $$(R.l=\mu)$$ of finite thickness $$t$$. The emerging light is

A
parallel to the incident ray

B
perpendicular to the incident ray

C
rotates by an angle $$\left[t \sin^{-1}\left( \dfrac {1}{\mu}\right)\right]$$

D
totally internally reflected

Solution

The emerging ray will be parallel to the incident ray.
Question 9
1 / -0

A light ray passes through a glass slab. Which of the following diagrams shows the correct path of the ray?

A

B

C

D

Solution

When light travels from a rarer to a denser medium, it bends towards the normal. And when light travels from denser to rarer medium, it bends away from the normal. Also, from the diagram, it is clear that emergent ray will always be parallel to the incident ray.
Question 10
1 / -0

A plane mirror is placed along the x-axis facing negative y-axis. The mirror is fixed. A point object is moving with $$3\hat{i} \, + \, 4\hat{j}$$ in front of the plane mirror. The relative velocity of image with respect to its object is

A
$$-8\hat{j}$$

B
$$8\hat{j}$$

C
$$3\hat{i} \, - \, 4\hat{j}$$

D
$$-6\hat{i}$$

Solution

Velocity of object, $$\bar{v}_{ob} = 3 \hat{i} + 4 \hat{j} $$
Velocity of image $$\bar{v}_{image} = 3 \hat{i} - 4 \hat{j}$$
Relative velocity of image with respect to its object $$\bar{v}_{rel} \, = \, \bar{v}_{image} \, - \, \bar{v}_{ob} \, = \, (3 \hat{i} \, - \, 4 \hat{j}) \, - \, (3 \hat{i} \, + \, 4 \hat{j}) = -8 \hat{j} $$