Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 71

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Question 1
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As shown in above figure, a beam of light in air is incident upon the smooth surface of glass $$30^o$$ with glass surface. Calculate the refractive index of the glass if the reflected beam and refracted beam are perpendicular to each other.

A
$$\frac{1}{2}$$

B
$$\frac{1}{2}\sqrt{3}$$

C
$$\sqrt{3}$$

D
$$2\sqrt{3}$$

E
$$3\sqrt{3}$$

Solution

From figure, $$i=90^{o}-30^{o}=60^{o}$$

And, from law of reflection , angle of reflection=$$r=i=60^{o}$$

Since, reflected and refracted ray are normal,
hence,
$$r+R+90^{o}=180^{o}$$ where $$R$$= angle of refraction
$$\implies R=90^{o}-r=90^{o}-60^{o}=30^{o}$$

From Snell's Law:-

$$\mu=\dfrac{\sin i}{\sin R}=\dfrac{\sin 60^{o}}{\sin 30^{o}}$$

$$\implies \mu=\sqrt{3}$$

Answer-(C)
Question 2
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An object is placed at 20 cm in front of a concave mirror produces three times magnified real image. What is focal length of the concave mirror?

A
15 cm

B
6.6 cm

C
10 cm

D
7.5 cm

Solution

Given : Object distance, $$u = -20$$ cm

A concave mirror forms a real magnified image only when the image to be formed is inverted i.e. $$m = -3$$

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

$$\therefore$$ $$-3 = \dfrac{-v}{-20}$$ $$\implies v = -60$$ cm

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

$$\therefore$$ $$\dfrac{1}{-60}+\dfrac{1}{-20} =\dfrac{1}{f}$$

$$\implies$$ $$f = \dfrac{-60}{4} = -15$$ cm

Thus focal length of the mirror is 15 cm.
Question 3
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In above shown figure, an object is kept 20 cm from the concave mirror and formed the real image at the distance of 5 cm from the mirror. Calculate the focal length of the mirror.

A
2.5 cm

B
4 cm

C
25 cm

D
20 cm

E
40 cm

Solution

Given :
$$u = -20$$ cm
$$v = -5 $$ cm

Using Mirror equation

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

$$\therefore$$ $$\dfrac{1}{-20}+ \dfrac{1}{-5} =\dfrac{1}{f}$$ $$\implies f = -4$$ cm
Thus focal length of concave mirror is $$4$$ cm.
Question 4
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If I want an image larger than the object (using a convex lens), where must I place the object and where will the image be formed?

A
Object Beyond Focus $$F$$ , Image at Infinity

B
Object Beyond Centre of lens $$C$$, Image at Centre $$C$$

C
Object Between Focus $$F$$ and Centre of lens $$C$$, Image at Focus

D
Object Before Focus $$F$$, Image on same side as object beyond Focus

Solution

As seen in diagram to get an enlarged image object should be placed before focus $$F$$ and virtual, enlarged and erect image is formed beyond $$F$$ but on same side as object.
Question 5
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Below is a drawing made by a student that represents a light ray entering a piece of plastic and passing into the plastic. The student drew the original light ray to enter the plastic at a $$60.0$$ degree angle with the normal as pictured.
He was able to look through the plastic and line up a ruler with the original light ray he drew to measure the angle of the light ray in the plastic. This angle is labeled in the drawing. The student used his drawing to determine the index of refraction for the plastic.
What value did he determine?

A
$$1.66$$

B
$$0.602$$

C
$$1.47$$

D
$$0.68$$

E
$$1.38$$

Solution

Here

$$i=60^{\circ}$$
$$r=36.1^{\circ}$$
From
$$n_a\sin i= n_p\sin r $$
$$1\sin 60^{\circ} = n_p\sin 36.1^{\circ} $$
$$1\times \frac{\sqrt{3}}{2}=n_p\times\sin 36.1^{\circ} $$
$$\sin 36.1^{\circ}=0.59$$
$$1\times \frac{\sqrt{3}}{2}=n_p\times 0.59 $$
$$ n_p=1.47$$
Question 6
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Which of the following is Snell's law

A
$$n_1 sini = n_2 sinr$$

B
$$n_{1}/n_{2} = \sin r \times constant$$

C
$$n_{2}/n_{1} = \sin r/ \sin i$$

D
$$n_{2} \sin i = constant$$

Solution

According to Snell's Law:-

$$\dfrac{\sin i}{\sin r}=\dfrac{n_2}{n_1}$$ when ray of light is incident from medium-1

$$\implies n_1\sin i=n_2\sin r$$

Answer-(A)
Question 7
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A ray of light traveling in the air is incident on the plane of a transparent medium. The angle of the incident is 45$$^o$$ and that of refraction is 30$$^o$$. Find the refractive index of the medium.

A
2

B
$$\dfrac{1}{\sqrt 2}$$

C
$$2 \sqrt 2$$

D
$$\sqrt 2$$

Solution

Snell's Law of refraction gives the relation between the incident and refracted light at the interface of two media, which is given by $$\displaystyle \frac{\mu_2}{\mu_1} = \frac{\sin i}{\sin r}$$

The initial medium is air. Thus $$\mu_1 = 1$$
Given incidence angle $$i = 45^o$$ and refraction angle is $$30^o$$

By Snell's law, $$\mu_2 = \mu_1\times \displaystyle \frac{\sin i}{\sin r} = 1\times \frac{\sin 45^o}{\sin 30^o} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$$
Question 8
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A concave mirror of focal length 'f' produces an image 'n' times the size of the object. If the image is real then the d distance of the object from the mirror is

A
$$(n + 1 )f$$

B
$$\dfrac{(n+1) f}{n}$$

C
$$\dfrac{(n-1) f}{n}$$

D
$$(n-1)f$$

Solution

Since the image is real, it must be inverted and magnification ratio should be negative.
Mgnification $$ = \dfrac{-v}{u}$$
$$-n=\dfrac{-v}{d}$$ , $$u=d$$ is the distance of object and and image distance is $$v$$
$$v=nd$$
Now using mirror formula
$$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$$
$$\dfrac{1}{f}=\dfrac{1}{nd}+\dfrac{1}{d}$$
So $$d=\dfrac{f(n+1)}{n}$$
Question 9
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A light is travelling from air into a medium. Velocity of light in a medium is reduced to $$0.75$$ times the velocity in air. Assume that angle of incidence '$$i$$' is very small, the deviation of the ray is

A
$$i$$

B
$$\dfrac { i }{ 3 } $$

C
$$\dfrac { i }{ 4 } $$

D
$$\dfrac { 3i }{ 4 } $$

Solution

Let speed of light in air be $$c$$.
Thus speed of light in medium $$v = \dfrac{3}{4}c$$
Using $$c\sin r = v\sin i$$
As $$i$$ and $$r$$ are very very small.
$$\therefore$$ $$c \times r = \dfrac{3}{4}c\times i$$
We get $$r = \dfrac{3i}{4}$$
Deviation of the ray $$\delta = i- r$$
$$\implies$$ $$\delta = i-\dfrac{3i}{4} = \dfrac{i}{4}$$
Question 10
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Study the given ray diagrams and select the correct statement from the following:

A
Device X is a concave mirror and device Y is a convex lens, whose focal lengths are 20cm and 25cm respectively.

B
Device X is a convex lens and device Y is a concave mirror, whose focal lengths are 20cm and 25cm respectively.

C
Device X is a concave lens and device Y is a convex mirror, whose focal lengths are 20cm and 25cm respectively.

D
Device X is a concave lens and device Y is a concave mirror, whose focal lengths are 20cm and 25cm respectively.

Solution

In the first case, The ray of light is converging, when passing from lens X, so it is a convex lens.
In the second case, the ray of light is diverging, when getting reflected from the mirror Y, so it is a concave mirror.