Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 37

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Question 1
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Ram and Shyam observe a virtual image formed by a real object placed a distance $$\dfrac{f}{2}$$ in front of a convex lens of focal length $$f$$. Now they want to use a concave lens having a focal length f so that a virtual image is formed at the same distance from the lens as before. Where should the object be kept relative to the concave lens

A
The object should be at innity

B
Object should be at a distance of $$f$$

C
Object should be at $$f/2$$

D
Object should be at $$2f$$

Solution

$$v= -\dfrac{2}{f}$$ ; $$f= -f$$

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

$$\dfrac{-2}{f}-\dfrac{1}{u}=\dfrac{-1}{f}$$

$$u= -f$$

It should be placed at a distance $$f$$ from the lens
option $$B$$ is correct
Question 2
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The cause of mirage formation in desert areas is

A
The refractive index of atmosphere decreases with decrease in height

B
The refractive index of atmosphere increases with decrease in height

C
The refractive index of atmosphere does not change with height

D
Scattering

Solution

The reason for mirage formation is "The refractive index of atmosphere increases with decrease in height" and hence the rays get totally refracted and forms a mirage.

As we go down in the atmosphere, the air gets denser and hence the velocity in that medium decreases. Since refractive index $$\mu = \dfrac{c}{v}$$. Hence, the refractive index increases as we decrease height.
Question 3
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If the light moving in a straight line bends by small but fixed angle while entering in different medium , it may be a case of :

A
reflection of light

B
refraction of light

C
scattering of light

D
dispersion of light

Solution

When light travels from one medium to another medium it changes direction and gets bend ,this process is called as refraction of light.It also changes the speed of light in the medium. When light travels from rarer medium to denser medium in bend towards the normal and vice- verca.
Hence it is a case of refraction of light.
Question 4
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Light is incident on a glass block as shown in Fig. If $$\theta_{1}$$ is increased slightly, what happens to $$\theta_{2}$$?

A
$$\theta_{2}$$ also increases slightly

B
$$\theta_{2}$$ is unchanged

C
$$\theta_{2}$$ decreases slightly

D
$$\theta_{2}$$ changes abruptly, since the ray experience total internal reflection
Question 5
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A ray of light is incident on a medium with angle of incidence $$i$$ and refracted into a second medium with angle of refraction $$r$$. The graph of $$sin(i)\ vs\ sin(r)$$ is as shown in fig. Then, the velocity of light in the first medium in $$n$$ times the velocity of light in the second medium. What should be the value of $$n$$?

A
$$\sqrt {3}$$

B
$$1/\sqrt {3}$$

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

D
$$2/\sqrt {3}$$

Solution

Using Snell's law, $$\dfrac{\mu_{2}}{\mu_{1}}=\dfrac{sini}{sinr}=tan60^o=\sqrt{3}$$

We know, $$\dfrac{v_{1}}{v_{2}}=\dfrac{\mu_{2}}{\mu_{1}}=n=\sqrt{3}$$

option $$A$$ is correct
Question 6
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Directions For Questions

A beam of light converges towards a point $$O$$, behind a convex mirror of focal length $$20\ cm$$.

...view full instructions

Find the magnification and nature of the image when point $$O$$ is $$30\ cm$$ behind the mirror.

A
$$2$$ (Virtual, Inverted)

B
$$3$$ (Real, Inverted)

C
$$5$$ (Real, Erect)

D
$$2$$ (Virtual, Erect)

Solution

For this situation, object will be virtual as shown in the figure.
Here, $$u = +10\ cm$$ and
$$f = +20\ cm$$.
$$\dfrac {1}{v} + \dfrac {1}{u} = \dfrac {1}{f}$$
So, $$\dfrac {1}{v} + \dfrac {1}{+10} = \dfrac {1}{+20} \Rightarrow i.e., v = -20\ cm$$
i.e., the image will be at a distance of $$20\ cm$$ in front of the mirror and will be real, erect and enlarged with
$$m=\frac{-v}{u}$$
$$m = \frac{-(-20)}{10} = +2$$.
(b.) For this situation also, object will be virtual as shown in the figure.
Here, $$u = +30\ cm$$
And $$f = +20\ cm$$
$$\dfrac {1}{v} + \dfrac {1}{+30} = \dfrac {1}{+20} i.e., v = +60\ cm$$
i.e., the image will be at a distance of $$60\ cm$$ behind the mirror and will be virtual, inverted, and enlarged with $$m = (-60/30) = -2$$.
Question 7
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A beam of light converges towards a point O, behind a convex mirror of focal length 20 cm. Find the magnification and nature of the image when point O is 30 cm behind the mirror.

A
2 (virtual, inverted)

B
3 (real, inverted)

C
3, (virtual, enlarged)

D
+1 (real, enlarged)

Solution

$$u=30$$ ; $$f=20$$

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

$$\dfrac{1}{v}+\dfrac{1}{30}=\dfrac{1}{20}$$

$$v= 60$$

Image is virtual (v>0)

Magnification is $$-\dfrac{v}{u}= -\dfrac{60}{30}= -2$$ (<0) hence it is inverted.
Question 8
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The diagram alongside shows the refraction of a ray of light from air to a liquid. Angle of incidence is :

A
$$30^o$$

B
$$45^o$$

C
$$90^o$$

D
$$60^o$$

Solution

Angle of incidence is the angle made between the incident ray and the normal. The angle between the incident ray and surface is $$30^o$$. Hence, angle between normal and incident ray will be 90-30=$$60^o$$
Question 9
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The diagram alongside shows the refraction of a ray of light from air to a liquid. Angle of refraction is:

A
$$60^o$$

B
$$15^o$$

C
$$30^o$$

D
$$45^o$$

Solution

Angle of refraction is the angle made between the refracted ray and the normal. The angle between the refracted ray and surface is $$45^o$$. Hence, angle between normal and refracted ray will be 90-45=$$45^o$$
Question 10
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A converging beam of rays is incident on a diverging thin lens. Having passed through the lens the ray intersect at a point 15 cm from the lens. If lens is removed the point where the rays meet will move 5 cm closer towards the mounting that holds the lens. The focal length of lens is

A
$$f=10 cm$$

B
$$f=15 cm$$

C
$$f=30 cm$$

D
$$f=40 cm$$

Solution

$$v= +15$$ ; $$u= +10$$

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

$$\dfrac{1}{15}-\dfrac{1}{10}=\dfrac{1}{f}$$

$$f= -30$$
option $$C$$ is correct