Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 75

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Weekly Quiz Competition

Question 1
1 / -0

A ray of light is travelling through a medium of refractive index $$\sqrt { 2 }$$ with respect to air. When it is incident on the surface making an angle $${ 45 }^{ o }$$ with the surface, which of the following will take place?(Medium to which refracted ray travels is air)

A
angle of refraction will be $${ 45 }^{ o }$$.

B
angle of refraction will be $${ 90 }^{ o }$$

C
The ray will be reflected internally

D
None of the above

Solution
Question 2
1 / -0

Monochromatic light is refracted from air into the glass of refractive index $$ \mu $$. The ratio of the wavelength of incident and refracted waves is

A
1:$$ \mu $$

B
1:$$ \mu $$$$ ^{2} $$

C
$$ \mu $$ :1

D
1:1

Solution
Question 3
1 / -0

The refractive indices of glass and water are 3/ 2 and 4/ 3 respectively. The refractive index of glass with respect to water is

A
2

B
8/9

C
9/8

D
5/3

Solution

Given
$$n_{glass}=3/2$$
$$n_{water}=4/3$$
Now refractive index of glass with respect to water is given by
$$_{water}n_{glass}$$=$$\dfrac{n_{glass}}{n_{water}}$$=$$\dfrac{3/2}{4/3}$$=$$\dfrac{3\times 3}{4 \times 2}$$= $$\dfrac{9}{8}$$
Question 4
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A ray of light is incident on the surface of separating two transparent medium at an angle $$45^0$$ and is refracted in medium at an angle $$30^0$$. Velocity of light in the medium will be...

A
$$3.8 \times 10^8 \ m/s$$

B
$$3.38 \times 10^8 \ m/s$$

C
$$2.12 \times 10^8 \ m/s$$

D
$$1.5 \times 10^8 \ m/s$$

Solution
Question 5
1 / -0

A glass containing a liquid appears to be half filled when viewed from top, if it is actually two-thirds filled. The refractive index of liquid is

A
$$\cfrac 7 6$$

B
$$\cfrac 5 4$$

C
$$\cfrac 4 3$$

D
$$2$$

Solution
Question 6
1 / -0

The image formed by a convex mirror of focal length 30 cm is a quarter of the size of the object. The distance of the object from the mirror is-

A
$$30cm$$

B
$$90 cm$$

C
$$120 cm$$

D
$$60 cm$$

Solution

Focal length of the convex mirror (f) = 30 cm
$${h_i} = \frac{{{h_0}}}{4}$$
$$ \Rightarrow m=\frac{{{h_i}}}{{{h_0}}} = \frac{1}{4} = - \frac{v}{u}=m$$
$$ \Rightarrow v = - \frac{u}{4}$$
The mirror formula is given as
$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$
Substituting the values in the formula,
$$\frac{1}{{30}} = \frac{1}{u} - \frac{4}{u}$$
$$\frac{1}{{30}} = - \frac{3}{u}$$
$$u = - 90{\text{ cm}}$$
Therefore, the distance of the object from the mirror is 90 cm.
Question 7
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A convex lens is placed somewhere in between an object and a screen. The distance between the object and screen is $$48\ cm$$. If the numerical value of the magnification produced by the lens is $$3$$, the focal length of the lens is :

A
$$16\ cm$$

B
$$4.5\ cm$$

C
$$12\ cm$$

D
$$9\ cm$$

Solution

The distance between object and screen $$ = 48\ cm $$
$$u + v = 48\ cm $$ ...............(i)

Magnification $$ = 3 $$
$$ \dfrac{v}{u} = 3 $$

$$\boxed{v = 3u} $$

Putting the value in equation (i)
$$u + 3u = 48$$
$$\boxed{u = 12\ cm} $$ $$\boxed{v = 3 \times 12 = 36 cm} $$

Focal length is given by
$$ \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} $$

$$ = \dfrac{1}{36} - \dfrac{1}{-12} $$

$$ \dfrac{1}{f} = \dfrac{1 + 3}{36} $$

$$\boxed{f = 9\ cm} $$
Question 8
1 / -0

A ray of light is incident on a glass at an angle of $$60^o$$. What would be the refractive index of glass, if reflected and refracted rays are perpendicular to each other ?

A
$$\sqrt{3}$$

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

C
$$\dfrac{\sqrt{3}}{2}$$

D
$$\dfrac{1}{2}$$

Solution

The incident ray $$1$$ will be reflected at an equal angle as reflected ray $$2.$$
By laws of reflection, $$i=r=60^o$$
The refracted ray $$3$$ is perpendicular to reflected ray $$2.$$
So by geometry, the angle of refraction $$r'=30^o$$
Now by Snell's law,
$$\mu_{air} \ sin \ i= \mu_{glass} \ sin \ r'$$
$$(1) \times sin \ 60^o= \mu_{glass} \times sin \ 30^o$$
$$\dfrac{\sqrt3}{2}=\mu_{glass} \times \dfrac{1}{2}$$
$$\mu_{glass}=\sqrt3$$
Option A is the answer.
Question 9
1 / -0

$${ n }_{ 12 }\times{ n }_{ 21 },$$ in simplest from is

A
1

B
2

C
0

D
4

Solution

Suppose $${\eta _1}$$ and $${\eta _2}$$ is the refractive index of medium $$1$$ and $$2$$.

The refractive index of the medium $$1$$ with respect to the medium $$2$$ is given as

$${\eta _{12}} = \dfrac{{{\eta _1}}}{{{\eta _2}}}$$

Similarly the refractive index of the medium $$2$$ with respect to the medium $$1$$ is given as

$${\eta _{21}} = \dfrac{{{\eta _2}}}{{{\eta _1}}}$$

Hence,

$${\eta _{12}} \times {\eta _{21}} = \dfrac{{{\eta _1}}}{{{\eta _2}}} \times \dfrac{{{\eta _2}}}{{{\eta _1}}}$$

$${\eta _{12}} \times {\eta _{21}} = 1$$.
Question 10
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Two objects A and B when placed in placed in turn in front of a concave mirror of focal length 7.5 cm give images of equal size. If A is three times the size of B and is placed 30 cm from the mirror, find the distance of B from the mirror.

A
-15 cm

B
25 cm

C
-17.5 cm

D
-11 cm

Solution

Given,

The size of the image of object A and B is the same, hence

$${h_A}^\prime = {h_B}^\prime $$

The size of object A is four times that of B, hence

$${h_A} = 3{h_B}$$

Now the magnification of object A can be given as

$${m_A} = \dfrac{{{h_A}^\prime }}{{{h_A}}}$$                                           …… (1)

Similarly for object B

$${m_B} = {\text{ }}\dfrac{{{h_B}^\prime }}{{{h_B}}}$$                                           ……. (2)

From equation (1) and (2) the ratio of magnification is given as

$$\dfrac{{{m_A}}}{{{m_B}}} = \dfrac{{{h_A}^\prime }}{{{h_A}}}{\text{ }} \times {\text{ }}\dfrac{{{h_B}}}{{{h_B}^\prime }} \Rightarrow \dfrac{{{m_A}}}{{{m_B}}} = \dfrac{{{h_A}^\prime }}{{3{h_B}}}{\text{ }} \times {\text{ }}\dfrac{{{h_B}}}{{{h_A}^\prime }}$$

$$\therefore \dfrac{{{m_A}}}{{{m_B}}} = \dfrac{1}{3}$$                                       ……. (3)

Now the magnification formula is given as

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

Therefore equation (3) becomes

$$\dfrac{{{m_A}}}{{{m_B}}} = \dfrac{{f - {u_B}}}{{f - {u_A}}} \Rightarrow \dfrac{1}{3} = \dfrac{{ - 7.5 - {u_B}}}{{ - 7.5 + 30}}$$

$$ - {u_B} = \dfrac{{22.5}}{3} + 10 \Rightarrow {u_B} = - 17.5cm$$

Hence the object B is placed at $$-17.5\;cm$$ from the mirror.