Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following is correct regarding reflection.

A
Reflection is a phenomenon by which light incident on a surface is sent back into the same medium

B
Reflection is a phenomenon by which light incident on a surface passes into another medium

C
Reflection is a phenomenon by which light incident on a surface is scattered

D
Reflection is a phenomenon by which light incident on a surface is absorbed by the surface

Solution

Reflection is a phenomenon in which the incident rays are sent back into same medium following laws of reflection.

Answer-(A)
Question 2
1 / -0

In optical instruments, the lenses are used to form the image by the phenomenon of ...................................

A
Reflection

B
Refraction

C
Scattering

D
Diffusion

Solution

In optical instruments, the lenses are used to form the images by the phenomenon of Refraction.
Question 3
1 / -0

Convex lens

A
Shrinks the image

B
Magnifies the image

C
Sharpens the image

D
Increase the contrast of the image

Solution

When the object is between the Focal point and the center of curvature then the image is formed beyond the imaginary point and behind the object which is virtual and magnified.
Question 4
1 / -0

A convex mirror of radius of curvature $$1.6 m$$ has an object placed at a distance of $$1m$$ from it. The image is formed at a distance of :

A
$$\dfrac{18}{3}m$$ behind the mirror

B
$$\dfrac{18}{3}m$$ infront of the mirror

C
$$\dfrac{4}{9}m$$ behind the mirror

D
$$\dfrac{4}{9}m$$ infront of the mirror

Solution

Given, $$f=\dfrac{1.6}{2}=0.8m$$ and $$u=-1m$$
$$\dfrac{1}{f} = \dfrac{1}{v}-\dfrac{1}{u}$$
$$\dfrac{1}{v} = \dfrac{1}{0.8} - \dfrac{1}{-1} = \dfrac{10}{8} + 1=\dfrac{9}{4}$$
$$\Rightarrow v=\dfrac{4}{9}m$$
Question 5
1 / -0

The apparent flattening of the sun at sunset and sunrise is due to

A
refraction

B
diffraction

C
total internal reflection

D
interference

E
polarization

Solution

The apparent flattering of the sun at sunset and sunrise is due to refraction, because the sun becomes visible a little before actual sunrise and remains visible a little after actual sunset. because of atmosphere refraction of light.
Question 6
1 / -0

If the angle of incidence is i and that of refraction is r. Then the speed of light in the medium to which the light is reflected from air is?

A
$$\displaystyle v=C\frac{\sin i}{\cos r}$$

B
$$\displaystyle v=C\frac{\cos r}{\cos i}$$

C
$$\displaystyle v=C\frac{\sin r}{\sin i}$$

D
$$\displaystyle v=C\frac{\sin i}{\sin r}$$

Solution

Given,
Angle of incidence is$$ i$$
Angle refraction is $$r$$

From Snell's law, refracting index of second medium with respect to first medium

$$\mu=\displaystyle\frac{\sin i}{\sin r}$$ ......(1)

also

$$\mu=\displaystyle\frac{c}{v}$$.........(2)

From above two equations
$$\therefore \displaystyle \mu=\frac{c}{v}=\frac{\sin i}{\sin r}$$

$$\Rightarrow v=c\cdot \left(\displaystyle\frac{\sin r}{\sin i}\right)$$.

Hence option C
Question 7
1 / -0

Lens spectacles are made from

A
Soft Glass

B
Hard Glass

C
Pyrex Glass

D
Flint Glass

Solution

Answer: Flint Glass.
Flint glass is any highly refractive glass used to make lenses and prisms. Because it absorbs most ultraviolet light but comparatively little visible light, it is also used for telescope lenses.
Question 8
1 / -0

If the angle of incidence is twice the angle of refraction in a medium of refractive index $$\mu$$, then the angle of incidence is :

A
$$2\, cos ^{-1}\dfrac{\mu}{2}$$

B
$$2\, sin ^{-1}\dfrac{\mu}{2}$$

C
$$2\, cos ^{-1} \mu$$

D
$$2\, sin ^{-1} \mu$$

Solution
Question 9
1 / -0

A ray of light is travelling through a medium of refractive index $$\dfrac {1}{\sqrt {2}}$$ with respect to air. When it is incident on the surface making an angle $$45^{\circ}$$ with the surface, which of the following will take place?

A
Angle of refraction will be $$45^{\circ}$$

B
Angle of refraction will be $$90^{\circ}$$

C
The ray will be internally reflected

D
None of the above

Solution

Given, $$\mu = \dfrac {1}{\sqrt {2}} w.r.t.$$ air $$i = 45^{\circ}, r = ?$$
Also, $$\mu = \dfrac {\sin i}{\sin r}$$
$$\Rightarrow \dfrac {1}{\sqrt {2}} = \dfrac {\sin 45^{\circ}}{\sin r} \Rightarrow \dfrac {1}{\sqrt {2}} =\dfrac {1/\sqrt {2}}{\sin r}$$
$$\Rightarrow \sin r = 1$$
$$\Rightarrow r = 90^{\circ}$$
Note that the ray is incident from denser medium and is refracted into a rarer medium air.
Question 10
1 / -0

How will you design a shaving mirror assuming that a person keeps it $$10cm$$ from his face and view the magnified image of the face at the closest comfortable distance of $$25cm$$. The radius of curvature of the mirror would then be:

A
$$24cm$$

B
$$-33.3cm$$

C
$$-24cm$$

D
$$30cm$$

Solution

Given:
A person keeps a mirror at a distance of $$10cm$$ from his face and view the magnified image of the face at the closest comfortable distance of$$ 25cm.$$
Since magnified image on other side of mirror is produced it is clearly a concave mirror with case of object between POLE and FOCUS of mirror.
To find the radius of curvature of the mirror
Solution:
From given criteria,
Object distance, $$u=-10cm$$
Image distance, $$v=25cm$$ refer above fig.
Applying the mirror formula, we get
$$\dfrac 1f=\dfrac 1v+\dfrac 1u\\\implies \dfrac 1f=\dfrac 1{25}+\dfrac 1{(-10)}\\\implies \dfrac 1f=\dfrac {-3}{50}\\\implies f=\frac{-50}{3}cm$$
And Radius of curvature, $$R=2f=2\times \frac{-50}{3}=-33.3cm$$
Hence the radius of curvature of the mirror will be $$-33.3 cm.$$

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Light Reflection and Refraction Test - 61

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Light Reflection and Refraction Test - 61

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Question 1

Reflection is a phenomenon by which light incident on a surface is sent back into the same medium

Reflection is a phenomenon by which light incident on a surface passes into another medium

Reflection is a phenomenon by which light incident on a surface is scattered

Reflection is a phenomenon by which light incident on a surface is absorbed by the surface

Solution

Question 2

Reflection

Refraction

Scattering

Diffusion

Solution

Question 3

Shrinks the image

Magnifies the image

Sharpens the image

Increase the contrast of the image

Solution

Question 4

$$\dfrac{18}{3}m$$ behind the mirror

$$\dfrac{18}{3}m$$ infront of the mirror

$$\dfrac{4}{9}m$$ behind the mirror

$$\dfrac{4}{9}m$$ infront of the mirror

Solution

Question 5

refraction

diffraction

total internal reflection

interference

polarization

Solution

Question 6

$$\displaystyle v=C\frac{\sin i}{\cos r}$$

$$\displaystyle v=C\frac{\cos r}{\cos i}$$

$$\displaystyle v=C\frac{\sin r}{\sin i}$$

$$\displaystyle v=C\frac{\sin i}{\sin r}$$

Solution

Question 7

Soft Glass

Hard Glass

Pyrex Glass

Flint Glass

Solution

Question 8

$$2\, cos ^{-1}\dfrac{\mu}{2}$$

$$2\, sin ^{-1}\dfrac{\mu}{2}$$

$$2\, cos ^{-1} \mu$$

$$2\, sin ^{-1} \mu$$

Solution

Question 9

Angle of refraction will be $$45^{\circ}$$

Angle of refraction will be $$90^{\circ}$$

The ray will be internally reflected

None of the above

Solution

Question 10

$$24cm$$

$$-33.3cm$$

$$-24cm$$

$$30cm$$

Solution

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