Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 76

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Question 1
1 / -0

The refractive index of material of a prism of angles $$45^{\circ}, -45^{\circ}$$, and $$-90^{\circ}$$ is $$1.5$$. The path of the ray of light incident normally on the hypotenuse side is shown in

A

B

C

D
Question 2
1 / -0

Sunlight that bounces off a surface is said to be __________ from the surface.

A
Radiated

B
Absorbed

C
Emitted

D
Reflected

Solution
Question 3
1 / -0

Refractive index of a rectangular glass slab is $$\mu = \sqrt { 3 }.$$ A light ray incident at an angle $$60 ^ { \circ }$$ is displaced laterally through $$2.5 { cm } .$$ Distance travelled by light in the slab is

A
$$4 cm$$

B
$$5 cm$$

C
$$2.5\sqrt { 3 } { cm }$$

D
$$3 cm$$

Solution

Given, R.I. of the glass slab $$\mu=\sqrt3$$,
Angle of incidence at the first surface of the glass, $$\theta_i=60^\circ$$.
We know, lateral shift, $$d=t\dfrac{sin(\theta_i-\theta_e)}{cos\theta_e}=2.5cm$$ (given).
Again, from geometry, $$t=lcos\theta_e$$.
Here, $$l$$ be the distance travelled by light in the slab,
And, $$\theta_e$$ and $$t$$ be the angle of refraction at the first surface and depth of the slab.
So, we have, $$d=lsin(\theta_i-\theta_e)$$......... (1)
Now, using Snell's law,
$$sin\theta_i=\mu sin\theta_e$$
Or, $$sin60^\circ=\sqrt3 sin\theta_e$$
Or, $$sin\theta_e=\dfrac{1}{2}$$
Or, $$\theta_e=30^\circ$$
Now, from the equation (1),
$$2.5=l\times sin(60^\circ -30^\circ)=l\times sin30^\circ$$
Or, $$l=5cm$$
So, the correct option is (B).
Question 4
1 / -0

A thin glass (refractive index 1.5) lens has optical power of -5D in air.Its optical power in a liquid medium with refractive index 1.6 will be:

A
-1D

B
1 D

C
25 D

D
-25 D

Solution
Question 5
1 / -0

Light propagates $$2\ cm$$ distance in glass of refractive index $$1.5$$ in time $$t_0$$. In the same time $$t_0$$, light propagates a distance of $$2.25\ cm$$ in a medium. The refractive index of the medium is :

A
$$4/3$$

B
$$3/2$$

C
$$8/3$$

D
$$none\ of\ these$$

Solution

In medium-1 (refractive index $$n_1=1.5$$), light travels a distance $$2\ cm$$ in time $$t_0$$.
The speed of light in medium-1 is $$v_1 = \dfrac{2\times 10^{-2}\ m}{t_0\ sec}$$
In medium-2 (refractive index $$n_2$$), light travels a distance $$2.25\ cm$$ in time $$t_0$$.
The speed of light in medium-2 is $$v_2 = \dfrac{2.25\times 10^{-2}\ m}{t_0\ sec}$$
From the definition of refractive index, we can write,
$$\dfrac{n_1}{n_2}=\dfrac{v_2}{v_1}$$
$$\Rightarrow \dfrac{1.5}{n_2}= \dfrac{2.25\times 10^{-2}}{2\times 10^{-2}}$$
$$\Rightarrow n_2 = \dfrac{1.5\times 2\times 10^{-2}}{2\times 10^{-2}}= \dfrac{4}{3}$$
Question 6
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A convex mirror of focal length $$10cm$$ is shown in figure. A linear object $$AB=5cm$$ is placed along the optical axis. Point $$A$$ is at distance $$25cm$$ from the pole of mirror. The size of image of $$AB$$ is:

A
$$2.5cm$$

B
$$0.64cm$$

C
$$0.36cm$$

D
None of these

Solution
Question 7
1 / -0

When an object is placed at a distance of $$25 \ cm$$ from a mirror, the magnification is $$m_1$$. But the magnification becomes $$m_2$$, when the object is moved $$15 \ cm$$ farther away with respect to the earlier position. If $$\dfrac{m_1}{m_2}=4$$, then find the focal length of the mirror and what type of mirror it is?

A
$$20 \ cm$$ , convex

B
$$20 \ cm $$ , concave

C
$$10 \ cm$$ , convex

D
$$10 \ cm $$ , concave
Question 8
1 / -0

The second lens in this optical instrument can not :

A
Cause the light rays to focus closer than they would with the first lens, acting alone

B
Cause the light rays to focus farther away than they would with the first lens acting alone

C
Cause the light beam to diverge after refraction from it

D
Make the beam parallel

Solution

Concave lens is used to diverge the light after refraction. Hence it can not converge light after refraction.
Question 9
1 / -0

An object of size $$7.5\ cm$$ is placed in front of a convex mirror of radius of curvature $$25\ cm$$ at a distance of $$40\ cm.$$ The size of the image should be

A
$$2.3\ cm$$

B
$$1.7\ cm$$

C
$$1\ cm$$

D
$$0.8\ cm$$

Solution

By using $$\dfrac{I}{O}=\dfrac{f}{f-u}$$
$$\Rightarrow \dfrac{I}{+(7.5)}=\dfrac{(25/2)}{\left(\dfrac{25}{2}\right)-(-40)}\Rightarrow I=1.78\ cm$$
Question 10
1 / -0

The radius of curvature of a concave mirror is $$40\ cm$$ and the size of the real image is twice as that of an object, then the object distance is:

A
$$60\ cm$$

B
$$20\ cm$$

C
$$40\ cm$$

D
$$30\ cm$$

Solution

Given, Concave mirror
Radius of curvature, $$R = -40\ cm $$
Magnification, $$m = -2 $$ (for real image $$m$$ is negative)

Focal length, $$f = \dfrac{R}{2} = \dfrac{-40}{2} = -20\ cm $$

We know, $$m = \dfrac{-v}{u} $$

$$ -2 = \dfrac{-v}{u}$$

$$ \Rightarrow u = \dfrac{v}{2} $$

$$\Rightarrow v = 2u $$

Using mirror formula,
$$ \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} $$

$$\Rightarrow \dfrac{1}{-20} = \dfrac{1}{2u} + \dfrac{1}{u} = \dfrac{3}{2u} $$

$$\Rightarrow 2u = 3 \times (-20) $$

$$u = \dfrac{-60}{2} = -30\ cm $$

$$u = -30\ cm $$

The object is $$30\ cm $$ from the mirror.