Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 36

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Weekly Quiz Competition

Question 1
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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 the magnification obtained is equal to:

A
$$-2$$

B
$$-1.5$$

C
$$-3$$

D
$$-1$$

Solution

Given, Concave mirror
object distance, $$u = -60\ cm$$
focal length, $$f = -40\ cm$$
image distance, $$v$$
Using mirror formula,
$$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$$

$$\dfrac{1}{-40} = \dfrac{1}{-60} + \dfrac{1}{v}$$

$$\dfrac{1}{v} = \dfrac{1}{-40} + \dfrac{1}{60} = \dfrac{-3 + 2}{120}$$

$$v = -120\ cm$$

Magnification, $$m$$
$$m = \dfrac{-v}{u}$$
$$m= \dfrac{-\left(-120\right)}{-60}$$
$$m = -2$$
Question 2
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A lens which is thicker in the middle and thinner at the edges is called a :

A
Convex lens

B
Concave lens

C
Cylindrical lens

D
None of these

Solution

A lens which is thicker in the middle and thinner at the edges is called a convex lens.
Question 3
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The bending of light as it passes from one medium into another is commonly known as

A
Reflection

B
Refraction

C
Scattering

D
Dispersion

Solution

In refraction ray of light passes from one medium to another medium. The ray of light bends towards normal in the denser medium.
Question 4
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The constant ratio of the sine of angle of incidence to the sine of angle of refraction is also known as:

A
Snell's law

B
Optical density

C
Relative density

D
None of these

Solution

Ratio of the sine of angle of incidence to the sine of angle of refraction is refracting index of second medium with respect to first medium or Optical density of second medium with respect to first medium.
Question 5
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When light travels from one medium into another it suffers

A
Reflection

B
Refraction

C
Dispersion

D
None of these

Solution

When light travels from one medium into another it suffers refraction,
Question 6
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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 the figure. Then the velocity of light in the first medium is ................. times the velocity of light in the second medium.

A
$$\sqrt{3}$$

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

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

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

Solution

Given
Ray of light is refracted when it is passed from one medium to another
Solution
Let refractive index of medium 1 is $$\mu_{1}$$
similarly refractive index of medium 2 is $$\mu_{2}$$
Using Snell law
$$\mu_{1}\sin i =\mu_{2}\sin r$$
$$\dfrac{\mu_{1}}{\mu_{2}}=\dfrac{\sin r}{\sin i}$$
Slope of graph$$=\tan 30 =\dfrac{\sin r}{\sin i}$$
$$\dfrac{\mu_{1}}{\mu_{2}}=3^{-1/2}$$

We know that velocity of light in a medium is inversely proportional to refractive index, ie $$\mu=\dfrac {c}{v}$$
$$\dfrac{\mu_{1}}{\mu_{2}}=\dfrac{v_{2}}{v_{1}}=3^{-1/2}$$

$$v_{1}=3^{1/2} \times v_{2}$$

The correct option is A
Question 7
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........ images cannot be caught on a screen.

A
Real

B
Inverted

C
Virtual

D
Erect

Solution

Virtual images cannot be caught on a screen as they are not formed due to actual intersection of rays coming from the object.
Question 8
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The expression for the magnification of a spherical mirror in the terms of focal length (f) and the distance of the object from mirror (u) is

A
$$\frac{-f}{u-f}$$

B
$$\frac{f}{u+f}$$

C
$$\frac{-f}{u+f}$$

D
$$\frac{f}{u-f}$$

Solution

Equation of spherical mirror is $$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$$ , where v is the image distance.
solving,
Replacing $$v$$ with $$v=-mu$$ ,
$$\dfrac{1}{f} = \dfrac{1}{u} (1 - \dfrac{1}{m} ) $$
$$u = f (\dfrac{1}{m} +1)$$ where m = magnification $$=- \dfrac{v}{u}$$
$$u =\dfrac{ f}{m} + f$$
$$m =- \dfrac{f }{ (u - f)}$$
Question 9
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Directions For Questions

This question concerns a symmetrical lens shown, along with its two focal points. It is made of plastic with ($$n=1.2$$), and has a focal length f. Four different regions are shown
Here (a) $$-\infty < x < - f$$, (b) $$-f < x < 0$$, (c) $$0 < x < f$$ (d) $$f < x < \infty$$

...view full instructions

If incident rays are converging then in which region does the image appear?

A
(a)

B
(b)

C
(c)

D
(d)

Solution

When incident rays coming parallel to the principle axis , strike the convex lens, they converge at the focus of the lens.

In the given case, the incident rays are converging in nature, hence, after refraction from a convex lens, which is made up of an optically denser material ($$n=1.2$$), they will converge at a point behind the focus of the lens.

Hence, the incident rays converge in region C) $$0<x<f$$
Question 10
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A convex lens forms an image of an object placed $$20 cm$$ away from it at a distance of $$20 cm$$ on the other side of the lens. If the object moves $$5 cm$$ towards the lens, then the image will be :

A
$$5 cm$$ toward the lens

B
$$5 cm$$ away from the lens

C
$$10 cm$$ toward the lens

D
$$10 cm$$ away from the lens

Solution

Given, $$u= -20 \space cm$$ ; $$v= +20 \space cm$$

From lens formula, $$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$$

$$\dfrac{1}{20}-\dfrac{-1}{20}=\dfrac{1}{f}$$

$$f= +10 \space cm$$

Now object is moved by $$5 \ cm$$ towards the lens
So now, $$u= -15 \space cm$$ ; $$f= +10 \space cm$$

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

$$\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{10}$$

$$v= +30 \space cm$$

So the image has moved $$(30-20)=10 \space cm$$
Image is moved $$10 \ cm$$ away from the lens.
option $$D$$ is correct