Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 69

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Question 1
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In a vessel, as shown in fig, point P is just visible when no liquid is filled in vessel through a telescope in the air. When liquid is filled in the vessel completely, point Q is visible without moving the vessel or telescope. Find the refractive index of the liquid.

A
$$\dfrac {\sqrt {14}}{3}$$

B
$$\dfrac {\sqrt {85}}{5}$$

C
$$\sqrt 2$$

D
$$\sqrt 3$$

Solution

When liquid is not filled in the vessel, point P can just be seen.
Thus the triangles, ABP and BCD are similar.
Thus $$\dfrac{PA}{AB}=\dfrac{BC}{CD}$$
Hence, $$CD=4R$$
Now after filling in a liquid of refractive index $$\mu$$, point Q can be seen. Hence the angle of refraction at D is $$r$$ as shown in the figure.
Applying snell's law to the refraction at point D,
$$\sin i=\mu \sin r$$
$$\dfrac{1}{\sqrt{5}}=\mu \dfrac{1}{\sqrt{17}}$$
Hence, $$\mu=\dfrac{\sqrt{85}}{5}$$
Question 2
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When light falls on a given plate at angle of incidence of $$60^o$$, the reflected and refracted rays are found to be normal to each other. The refractive index of the material of the plate is then

A
0.866

B
1.5

C
1.732

D
2

Solution

Given:
Angle of incidence $$i = 60^o$$.
$$i+r = 90^o$$
$$\therefore r = 30^o$$
From Snell's law of refraction we can write,
Now $$\mu = \displaystyle \frac{sin i}{sin r} = \frac{sin 60}{sin 30} = \frac{\sqrt 3/2}{1/2} = \sqrt{3} = 1.732$$
Question 3
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Which of the following correctly represents the graphical variation between very small angles of incidence (i) and refraction (r) ?

A

B

C

D

Solution

We know that $$\dfrac{\sin{i}}{\sin{r}}=\mu$$

For small $$i$$ and $$r$$, $$\sin{i}\approx i$$ and $$\sin{r}\approx r$$

$$\implies \dfrac{i}{r}=\mu$$

$$\implies i=\mu r$$

Hence, the graph will be a straight line passing through the origin.

Hence, the answer is option-(D).
Question 4
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A real object is placed at a distance $$\dfrac{f}{2}$$ in front of a concave mirror having focal length $$f$$. If object is shifted by a distance $$\dfrac{f}{4}$$ away from the mirror. Find the ratio of final magnification to initial magnification.

A
$$4$$

B
$$\dfrac{1}{4}$$

C
$$\dfrac{1}{2}$$

D
$$2$$

Solution

Initially u$$=-\dfrac{f}{2}$$ ; $$ f= -f$$
$$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$$
or, $$\dfrac{1}{v}-\dfrac{2}{f}= -\dfrac{1}{f}$$
or, $$\dfrac{1}{v}=\dfrac{2}{f}-\dfrac{1}{f}$$
or, $$\dfrac{1}{v}=\dfrac{1}{f}$$
or, $$v=f$$
Initial magnification is $$-\dfrac {v}{u}=\dfrac{f}{(f/2)}$$= 2
After displacing $$u= - \dfrac{3f}{4}; f= -f $$
So, $$\dfrac{1}{v}-\dfrac{4}{3f}= -\dfrac{1}{f}$$
or, $$v=3f$$
Final magnification is $$-\dfrac{v}{u}=\dfrac{3f}{(3f/4)}=4$$
Ratio of final to initial magnification is: $$\dfrac{4}{2}= 2$$
Question 5
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When a light ray moves from air to a transparent medium the angle of incidence and the angle of refraction are $$45^{\circ}$$ and $$30^{\circ}$$ respectively. Find the refractive index of this medium is

A
$$1.21$$

B
$$1.42$$

C
$$1.88$$

D
$$2.42$$

Solution

Given,
Angle of incidence $$\angle i={ 45 }^{ 0 }$$
Angle of refraction $$\angle r={ 30 }^{ 0 }$$

By Snell's law
$$n=\dfrac { sini }{ sinr } $$, where $$n$$ is the refractive index of the medium.

$$n=\dfrac { sin{ 45 }^{ 0 } }{ sin{ 30 }^{ 0 } } =\sqrt { 2 }=1.42 $$
Question 6
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The image of an object placed on the principal axis of a concave mirror of focal length $$12\ cm$$ is formed at a point which is $$10\ cm$$ more distant from the mirror than the object. The magnification of the image is

A
$$8/3$$

B
$$2.5$$

C
$$2$$

D
$$-1.5$$

Solution

Let the object distance be $$u$$ then image distance is $$u+10$$
$$u= -u$$ ; $$v= -(u+10)$$ ; $$f= -12$$
$$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$$
$$\dfrac{-1}{u+10}+\dfrac{-1}{u}=\dfrac{-1}{12}$$
$$\dfrac{2u+10}{u(u+10)}=\dfrac{1}{12}$$
$$u=20$$cm
$$v=-30$$cm
Magnification is $$-\dfrac{v}{u}= -\dfrac{30}{20}= -1.5$$
Question 7
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The refractive indices of substances P,Q, Rand S are 1.38, 1.46, 1.56 and 1.24 respectively. When light travelling in air is incident on these substances at equal angles, the angle of refraction will be maximum in

A
P

B
Q

C
R

D
S

Solution

The angle of refraction($$r$$) will be the maximum when the light bends the least on changing the medium. This will happen when the refraction index of the medium is the least.
Thus, the angle of refraction will be maximum in the medium of refractive index 1.24.
Question 8
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In the figure below, PQRS denotes the path followed by a ray of light as it travels three media in succession. The absolute refractive indices of the media are $$\mu_1 \, \mu_2\, and\, \mu_3 $$ respectively. (The line segment RS' in the figure is parallel to PQ). Then :

A
$$\mu_1\, > \, \mu_2\, >\, \mu_3$$

B
$$\mu_1\, < \, \mu_3\, \mu_2$$

C
$$\mu_1\, = \, \mu_3\, <\, \mu_2$$

D
$$\mu_1\, < \, \mu_3\, <\, \mu_2$$

Solution

From the given figure we can analyse $$\mu_2>\mu_1 ; \mu_3<\mu_2 ; \mu_1<\mu_3$$
so, $$\mu_1<\mu_3<\mu_2$$
Question 9
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An object is placed at a distance of 20 cm from a convex mirror or radius of curvature 40 cm. At what distance from the object should a plane mirror be placed so that the images due to the mirror and the plane mirror are in the same plane?

A
15 cm

B
30 cm

C
60 cm

D
40 cm

Solution

Given: An object is placed at a distance of 20 cm from a convex mirror or radius of curvature 40 cm.
To find the distance at which the plane mirror should be placed such that the images due to the mirror and the plane mirror are in the same plane
Solution:
According to the given criteria,
Object distance from convex mirror, $$u=-20cm$$
Radius of curvature of convex mirror, $$R=40cm$$
Focal length of the convex mirror, $$f=\dfrac R2=\dfrac {40}2=20cm$$
Now applying the mirror formula we get,
$$\dfrac 1f=\dfrac 1v+\dfrac 1u\\\implies \dfrac 1{20}=\dfrac 1v+\dfrac 1{-20}\\\implies \dfrac 1v=\dfrac 1{20}+\dfrac 1{20}\\\implies v=10cm$$
So the distance is formed on the opposite side of the object at a distance of 10 cm from the mirror as shown in above fig.
The distance between object and image is $$u+v=20+10=30cm$$
When plane mirror is placed in front of object and in order to get image at same position as convex mirror, the mirror should be place at
$$\dfrac {30}2=15cm$$, as in plane mirror the object distance from mirror is equal to image distance from the plane mirror (refer above fig).
Hence the distance at which the plane mirror should be placed such that the images due to the mirror and the plane mirror are in the same plane = 15cm
Question 10
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An object is placed on the principal axis of a concave mirror at a distance of $$60\ cm$$. If the focal length of the concave mirror is $$40\ cm$$ then determine the magnification of the obtained image.

A
$$4$$

B
$$-2$$

C
$$-4$$

D
$$+2$$

Solution

Given:
Distance of object $$u=-60\ cm$$
Focal length $$f=-40\ cm$$
From mirror formula
$$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$$
$$\dfrac{1}{v}-\dfrac{1}{60}=\dfrac{-1}{40}$$
$$\dfrac{1}{v}=\dfrac{1}{60}-\dfrac{1}{40}$$
$$v=-120 cm$$
Hence magnification is given by
$$m=\dfrac{-v}{u}=-\dfrac{(-120)}{(-60)}$$
$$m=-2$$