Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 65

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Weekly Quiz Competition

Question 1
1 / -0

An object is placed first at infinity and then at $$20\ cm$$ from the object side focal plane of the convex lens. The two images thus formed are $$5\ cm$$ apart. The focal length of the lens is:

A
$$5\ cm$$

B
$$10\ cm$$

C
$$15\ cm$$

D
$$20\ cm$$

Solution

Apply lens formula :
$$\dfrac{ 1 }{ f } = \dfrac{ 1 }{ v } - \dfrac{ 1 }{ u }$$
$$f$$: Focal length of the lens
$$v$$: Image distance from the lens
$$u$$: Object distance from the lens
For first case:
$$u = \infty$$
$$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{\infty}$$
$$ \Rightarrow f=v$$
For second case:
$$\dfrac{1}{f}=\dfrac{1}{(f+5)}-\dfrac{1}{-(f+20)}$$
$$ \Rightarrow f=10\ cm$$
Question 2
1 / -0

When a glass rod is immersed in a liquid of same refractive index, it will _____.

A
Appear bent

B
Disappear

C
Appear longer

D
Not be affected

Solution
Question 3
1 / -0

When a beam of light goes from a denser medium $$ \left(\mu_{d}\right) $$ to a rarer medium ( $$ \mu_{r} ), $$ then it is generally observed that magnitude of angle of incidence is half that of angle of refraction. Then magnitude of incident angle will be - (here $$ \mu=\mu_{d} / \mu_{r} ) $$

A
2$$ \sin ^{-1}\left(\frac{\mu}{2}\right) $$

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

C
$$ \cos ^{-1}\left(\frac{\mu}{2}\right) $$

D
2$$ \cos ^{-1}\left(\frac{\mu}{2}\right) $$
Question 4
1 / -0

At what distance a person should face a concave mirror of focal length $$0.4 \ m$$ so that magnification is 5 times for a virtual image

A
$$32 \ cm$$

B
$$24 \ cm$$

C
$$16 \ cm$$

D
$$50 \ cm$$

Solution

Given:
Focal length of the concave mirror $$f=-0.4 \ m=-40 \ cm$$
Magnification $$m=+5$$ (positive because the image is virtual)

Let us assume the object distance is $$d$$ (in cm).
So object distance $$u=-d$$

We know that for a spherical mirror
$$m=-\dfrac{v}{u}$$
$$\Rightarrow +5=-\dfrac{v}{-d}$$
$$\Rightarrow v=+5d$$ is the image distance.

By mirror formula,
$$\dfrac{1}{v}+ \dfrac{1}{u}=\dfrac{1}{f}$$
$$\Rightarrow \dfrac{1}{+5d}+\dfrac{1}{-d}=\dfrac{1}{-40}$$

$$\Rightarrow \dfrac{1-5}{5d}=- \dfrac{1}{40}$$

$$\Rightarrow\dfrac{5d}{4}=40$$
$$\Rightarrow d=32 \ cm$$

So, object distance $$u=-d=-(32 \ cm)=-32 \ cm$$
Question 5
1 / -0

A ray of light is incident on the surface of transparent medium at an angle of $$45^o$$ and is refracted in the medium at an angle of $$30^o$$. What will be the velocity of light in the transparent medium ?

A
$$1.96 \times 10^8 m/s $$

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

C
$$2.65 \times 10^8 m/s $$

D
$$1.25 \times 10^8 m/s $$

Solution

From Snell's Law, $$\mu = \dfrac{\sin i}{\sin r} = \dfrac{c}{v}$$

$$\Rightarrow v = c\dfrac{\sin r}{\sin i} = 3\times 10^8 \times \dfrac{\sin 30^o}{\sin 45^o}$$

$$\Rightarrow v=3\times 10^8\times \dfrac{\dfrac{1}{2}}{\dfrac{1}{\sqrt{2}}} = 2.12 \times 10^8 m/s$$
Question 6
1 / -0

An object is placed at the principal focus of a convex mirror. The image will be at?

A
Centre of curvature

B
Principal focus

C
Infinity

D
No image will be formed

Solution

No image is formed.

Light rays reflected from the mirror neither converge (forming a real image) nor diverge (allowing a virtual image to be seen).

Light rays from any point on the object are reflected so that they are parallel to each other. It’s simply the reverse case of parallel light rays hitting the mirror and forming an image at the principal focus.
When the object is placed at the focus the image is formed at infinity &highly enlarged.
Question 7
1 / -0

A Point source of Light is placed $$4 \ m$$ below the surface of water of refractive index $$\dfrac{5}{3}$$. the minimum diameter of a disc which should be placed over the source on the surface of water to cut-off all light coming out of water is:

A
$$2 \ m $$

B
$$ 6 \ m$$

C
$$4\ m$$

D
$$3\ m$$

Solution

Since the light is in the form of a disc, it implies, that the rays after refraction at the interface of the two media, must be going parallel to the surface.
Applying Snell's law at the interface of the two media, we have
μ1sini=μ2sinr
Given,
μ1=μliquid=53
We know, μ2=μair=1
From the figure,
i=θ and r=90o
⇒53sinθ=1
sinθ=35
From the figure,
sinθ=rr2+h2√
Given, h = 4 cm
rr2+h2√=35
Squaring both sides, we have
r2r2+42=925
r2+16r2=259
16r2=259−1⇒16r2=169
r2=16×916
r=3cm
Diameter, d=2×r=2×3=6cmm
Question 8
1 / -0

An object is placed at a distance of $$20$$cm from the pole of a concave mirror of focal length $$10$$cm. The distance of the image formed is?

A
$$+20$$ cm

B
$$+10$$ cm

C
$$-20$$ cm

D
$$-10$$ cm

Solution

$$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$$
$$\Rightarrow\dfrac{1}{v}+\dfrac{1}{-20}=\dfrac{-1}{10}$$
$$\Rightarrow \dfrac{1}{v}=-\dfrac{1}{10}+\dfrac{1}{20}=\dfrac{-2+1}{20}$$
$$=\dfrac{-1}{20}$$
$$v=-20cm$$
The object is placed at center of curvature & hence will him image at $$'c'.$$
Question 9
1 / -0

A candle placed $$25$$cm from a lens forms an image on a screen placed $$75$$cm on the other side of the lens. The focal length and type of the lens should be?

A
$$+18.75$$cm and convex lens

B
$$-18.75$$cm and concave lens

C
$$+20.25$$cm and convex lens

D
$$-20.25$$cm and concave lens

Solution

$$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\rightarrow \dfrac{1}{75}-\dfrac{1}{-25}=\dfrac{1}{f}$$
$$\Rightarrow \dfrac{1}{f}=\dfrac{1}{25}+\dfrac{3}{75}=\dfrac{4}{75}$$
$$f=18.75$$cm & his the
Hence the lens should be convex.
Question 10
1 / -0

Calculate the refractive index of glass with respect to water. It is given that refractive indices of glass and water with respect to air are $$\dfrac {3}{2}$$ and $$\dfrac {4}{3}$$ respectively.

A
$$\dfrac {8}{9}$$

B
$$\dfrac {9}{8}$$

C
$$\dfrac {7}{6}$$

D
$$none\ of\ these$$

Solution

Given:
The refractive index of glass with respect to air is $$n_g=\dfrac{3}{2}$$
The refractive index of water with respect to air is $$n_w=\dfrac{4}{3}$$

Let the refractive index of glass with respect to water is $$n_{gw}$$
According to the definition, $$n_{gw}=\dfrac{n_g}{n_w}=\dfrac{3\times 3}{2\times 4}=\dfrac{9}{8}$$

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Light Reflection and Refraction Test - 65

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Light Reflection and Refraction Test - 65

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SHARING IS CARING

Question 1

$$5\ cm$$

$$10\ cm$$

$$15\ cm$$

$$20\ cm$$

Solution

Question 2

Appear bent

Disappear

Appear longer

Not be affected

Solution

Question 3

2$$ \sin ^{-1}\left(\frac{\mu}{2}\right) $$

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

$$ \cos ^{-1}\left(\frac{\mu}{2}\right) $$

2$$ \cos ^{-1}\left(\frac{\mu}{2}\right) $$

Question 4

$$32 \ cm$$

$$24 \ cm$$

$$16 \ cm$$

$$50 \ cm$$

Solution

Question 5

$$1.96 \times 10^8 m/s $$

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

$$2.65 \times 10^8 m/s $$

$$1.25 \times 10^8 m/s $$

Solution

Question 6

Centre of curvature

Principal focus

Infinity

No image will be formed

Solution

Question 7

$$2 \ m $$

$$ 6 \ m$$

$$4\ m$$

$$3\ m$$

Solution

Question 8

$$+20$$ cm

$$+10$$ cm

$$-20$$ cm

$$-10$$ cm

Solution

Question 9

$$+18.75$$cm and convex lens

$$-18.75$$cm and concave lens

$$+20.25$$cm and convex lens

$$-20.25$$cm and concave lens

Solution

Question 10

$$\dfrac {8}{9}$$

$$\dfrac {9}{8}$$

$$\dfrac {7}{6}$$

$$none\ of\ these$$

Solution

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