Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 74

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Question 1
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A pin of length $$2\ cm$$ along the principle axis of a converging lens, the centre being at a distance of $$11\ cm$$ from the lens. The focal length of the lens is $$6\ cm$$. Find the size of the image ?

A
$$3\ cm$$

B
$$2.4\ cm$$

C
$$10\ cm$$

D
$$5\ cm$$

Solution

REF.Image.
Now we have to calculate the image of A and B. Let the images be A', B'. So, length
of A' B' = size of image.
For A, u = -10 cm, f = 6 cm
Since, $$\displaystyle \frac{1}{v}-\frac{1}{u} = \frac{1}{f}\Rightarrow \frac{1}{v}-\frac{1}{-10} = \frac{1}{6}$$
$$ \Rightarrow v = 15 cm = OA'$$
For B, u = -12 cm, f = 6 cm
Again, $$\displaystyle \frac{1}{v}-\frac{1}{u} = \frac{1}{f}\Rightarrow \frac{1}{v} = \frac{1}{6}-\frac{1}{12}$$
$$ \Rightarrow $$ v = 12 cm = OB'
$$ \therefore $$ A'B' = OA' - OB' = 15 - 12 = 3 cm.
So, size of image = 3 cm.
Question 2
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An object placed in front of a concave mirror of focal length $$0.15\ m$$ produces a virtual image, which is twice the size of the object. The position of the object with respect to the mirror is :

A
$$-5.5\ cm$$

B
$$-6.5\ cm$$

C
$$- 7.5\ cm$$

D
$$- 8.5\ cm$$

Solution

Given:
Focal length $$f=-0.15$$m (concave mirror)
Magnification $$m=2$$
Let the position of the object is $$=-u$$
Since the image is virtual, it is erect.
So, magnification $$m=-\dfrac{v}{u}=-\dfrac{v}{(-u)}=2$$
$$\Rightarrow v=2u$$
Now, by mirror formulae
$$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$$
$$\Rightarrow \dfrac{1}{2u}+\dfrac{1}{-u}=\dfrac{1}{-0.15}$$
$$\Rightarrow -\dfrac{1}{2u}=\dfrac{1}{-0.15}$$

$$\Rightarrow u=\dfrac{0.15}{2}$$m

$$\Rightarrow u=\dfrac{15}{2}$$cm$$=7.5$$cm
Sign convention has been already used, So $$u=-7.5cm$$
So, Ans is (C).
Question 3
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A ray of light travels from water to glass as shown below. The refractive index of water is $$1.3$$ and the refractive index of glass is $$1.5$$. What is the angle of refraction ?

A
$$30.7^{o}$$

B
$$35.3^{o}$$

C
$$41.7^{o}$$

D
$$48.6^{o}$$

Solution
Question 4
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Which mirror is to be used to obtain a parallel beam of light from a point source?

A
Plane mirror

B
Concave mirror

C
convex mirror

D
None of these.

Solution

A point source when kept at the focus of a concave mirror, gives a parallel beam of light after reflection.
Question 5
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A film projector magnifies a film of area $$100 $$ square centimeter on screen. If linear magnification is $$4$$ then area of magnified image on screen will be-

A
$$1600 sq. cm$$

B
$$800 sq. cm$$

C
$$400 sq. cm$$

D
$$200 sq. cm$$

Solution

As linear magnification, $$M=4$$
Hence, a real magnification $${ m }_{ r }={ m }^{ 2 }$$
$${ \left( 4 \right) }^{ 2 }=16$$
Surface area of film image on screen $$=16\times 100=1600$$ $${ cm }^{ 2 }$$.
Question 6
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An object is placed at a distance of $$16 cm$$ from a convex mirror of focal length $$20 cm$$, the position of image with its nature is

A
Virtual, erect behind the mirror at 8.9 cm

B
Virtual, erect behind the mirror at 89 cm

C
Real, erect behind the mirror at 8.9 cm

D
Virtual, inverted behind the mirror at 89 cm

Solution

Given:
Object distance $$u=-16 \mathrm{~cm}$$
Focal length $$f=+20 \mathrm{~cm}$$ (+ve for convex mirror)

By mirror formula
$$\Rightarrow \dfrac{1}{v}+\dfrac{1}{4}=\dfrac{1}{f}$$
$$\Rightarrow \dfrac{1}{v}-\dfrac{1}{16}=\dfrac{1}{20}$$
$$\Rightarrow \dfrac{1}{v}=\dfrac{1}{20}+\dfrac{1}{16}$$

$$\therefore \quad \dfrac{1}{v}=\dfrac{4+5}{80}$$

$$\therefore v=\dfrac{80}{9}=+8.9 c m$$

Magnification $$(m)=\dfrac{-v}{u}$$
$$=\dfrac{-8.9}{-16}$$

$$=+0.556$$

$$m=$$ +ve, therefore erect and virtual image is formed. And image is formed at $$8.9 \mathrm{~cm}$$ behind the mirror.
Hence, (A) is correct option.
Question 7
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The focal length of a concave mirror is $$50 \ cm$$. Where should an object be placed, so that its image is two times and is inverted

A
$$75 \ cm$$

B
$$60 \ cm$$

C
$$125 \ cm$$

D
$$50 \ cm$$

Solution

Magnification formula for a mirror is given as $$m=-\dfrac{v}{u}$$

Given, that the image formed should be two times and inverted
Hence, $$m=-2$$

$$\Rightarrow -2 = -\dfrac{v}{u}$$
$$\Rightarrow v=2u$$

Using the mirror's formula , $$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$$

Since, mirror is concave. We have $$f=-50 \ cm$$

$$\Rightarrow \dfrac{1}{2u} + \dfrac{1}{u} = - \dfrac{1}{50}$$

$$\Rightarrow \dfrac{3}{2u} = - \dfrac{1}{50}$$

$$\Rightarrow u=-75 \ cm$$

Hence, object should be placed $$75 \ cm$$ in front of the mirror.
Question 8
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A real object is placed at $$20 \ cm$$ from a concave mirror of focal length $$5 \ cm$$. Find the magnification.

A
$$3$$

B
$$-3$$

C
$$\cfrac{1}{3}$$

D
$$-\cfrac{1}{3}$$

Solution

Give:
Object distance $$u=-20 \ cm$$
Focal length $$f=-5 \ cm$$

By mirror formula,
$$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$$

$$\Rightarrow \dfrac{1}{(-5)}=\dfrac{1}{(-20)}+\dfrac{1}{v}$$

$$\Rightarrow \dfrac{1}{v}=\dfrac{1}{-5}+\dfrac{1}{20}=\dfrac{-4+1}{20}=\dfrac{-3}{20}$$

$$\Rightarrow v=-\dfrac{20}{3} \ cm$$

Magnification $$m$$ is:
$$m=-\dfrac{v}{u}=-\dfrac{\left( \dfrac{-20}{3}\right)}{-20}$$

$$\Rightarrow m = -\dfrac{20}{3 \times 20}=-\dfrac{1}{3}$$

Option $$(D)$$ is the answer.
Question 9
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The image formed by a convex mirror of focal length $$30\ cm$$ is a quarter of the size of the object. The distance of the object from the mirror is

A
$$30\ cm$$

B
$$90\ cm$$

C
$$120\ cm$$

D
$$60\ cm$$

Solution

$$m = \dfrac{{ - v}}{u}$$
$$\left| V \right| = \dfrac{{\left| u \right|}}{u}$$
using mirror formula
$$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$$
$$ = \dfrac{u}{u} + \dfrac{1}{{ - u}} = \dfrac{1}{{30}}$$
$$ = \dfrac{3}{4} = \dfrac{1}{{30}}$$
$$ = u = 90\,cm.$$
Hence$$,$$ option $$(B)$$ is correct$$.$$
Question 10
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For water $$\mu = \dfrac{4}{3}$$ and the velocity of light in vacuum is $$3 \times 10^8 m/s$$, the time taken for light to travel a distance of $$450\ m$$ in water will be.

A
$$2 \mu s$$

B
$$1.5 \mu s$$

C
$$1 \mu s$$

D
$$3 \mu s$$

Solution