Light Reflection and Refraction Test

Result

Light Reflection and Refraction Test - 43

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A convex mirror of focal length $$f$$ (in air) is immersed in a liquid $$\displaystyle \left( \mu =\frac { 4 }{ 3 } \right) $$. The focal length of the mirror in liquid will be:

A
$$\displaystyle \left( \frac { 3 }{ 4 } \right) f$$

B
$$\displaystyle \left( \frac { 4 }{ 3 } \right) f$$

C
$$\displaystyle f$$

D
$$\displaystyle \left( \frac { 7 }{ 3 } \right) f$$

Solution

Mirrors, spherical or otherwise are based on the laws of reflection. That behavior is not affected by the medium in which the light travels, so the mirror's focal length would be no different in the air or any other medium.
Question 2
1 / -0

An image of the same size as that of the object cannot be produced by :

A
a plane mirror

B
a concave mirror

C
a convex mirror

D
a convex lens

Solution
Question 3
1 / -0

A student uses a converging lens to produce an enlarged virtual image of a scale she wishes to read accurately. The focal length of the lens is $$10\;cm$$. What is a suitable distance between the scale and the lens ?

A
$$8\;cm$$

B
$$10\;cm$$

C
$$15\;cm$$

D
$$20\;cm$$

Solution

$$f=10cm$$
we know that virtual image formed by converging lens only when we keep object between f and lens.hence suitable distance between the scale and the lens should be <10cm which is 8cm.
Question 4
1 / -0

Which of the following does not represent correct refraction?

A

B

C

D

Solution

According to the law of refraction if a ray of light passes through rare medium to denser medium it bends towards the normal from its path and vice-versa. So, all the cases following the same, but in the option $$B$$, the ray passes through rare medium to denser medium and its bends away from normal.
Hence, option $$B$$ is correct.
Question 5
1 / -0

Magnification produced by a convex mirror is always:

A
equal to 1

B
less than 1

C
more than 1

D
zero

Solution

A convex mirror always creates a virtual image which is diminished. So, magnification produced by convex mirror is always less than one.
Question 6
1 / -0

A ray of light enters a slab of material with increasing refractive index. Four possibilities of the trajectory of the ray are shown below.
The correct choice is:

A
A

B
B

C
C

D
D

Solution

According to definition:The refraction of light when it passes from a rare to a denser medium bends the light ray toward the normal and vice versa,but case A,B,D does not follow the same but C does.
Hence, the correct choice is C.
Question 7
1 / -0

An under-water swimmer cannot see very clearly even in absolutely clear water because of:

A
Absorption of light in water

B
Scattering of light in water

C
Reduction of speed of light in water

D
Change in the focal length of eye lens.
Question 8
1 / -0

Linear magnification produced by a concave mirror may be

A
Equal to I

B
Less than I

C
More than I

D
All of the above

Solution

Answer is D.

Magnification is defined as $$ m = \dfrac{height \ of \ image}{height \ of \ object} $$

A concave mirror forms real, diminished images, virtual images, real enlarged images as well as real image of same size.
So, the magnification can be less than 1, more than 1 or equal to 1.

Hence, all the given options are correct.
Question 9
1 / -0

If linear magnification for a spherical mirror is $$\dfrac{3}{2}$$, then we may write: (symbols have their usual meanings)

A
$$f=\dfrac{u}{2}$$

B
$$f=\dfrac{3u}{2}$$

C
$$f=\dfrac{3u}{5}$$

D
None of these

Solution

Mirror equation is: $$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$$
Multiplying both sides by $$u$$, we get:
$$\dfrac{u}{f}=\dfrac{u}{v}+1$$
Or $$\dfrac{u}{v}=\dfrac{u}{f}-1=\dfrac{u-f}{f}$$
Or $$\dfrac{v}{u}=\dfrac{f}{u-f}$$
Now, magnification, $$m=\dfrac{v}{u}=\dfrac{3}{2}$$
$$\therefore$$ $$\dfrac{f}{u-f}=\dfrac{3}{2}$$
Solving the above equation we get $$\dfrac{5}{2}f=\dfrac{3}{2}u$$
$$\implies f=\dfrac{3u}{5}$$
Question 10
1 / -0

The linear magnification for a mirror is the ratio of the size of the image to the size of the object, and is denoted by $$m$$. Then we may write: (symbols have their usual meanings)

A
$$f=\dfrac{1+m}{mu}$$

B
$$f=mu$$

C
$$f=\dfrac{mu}{1+m}$$

D
None of these

Solution

Mirror equation: $$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$$
Multiplying both sides by $$u$$, we get:
$$\dfrac{u}{f}=\dfrac{u}{v}+\dfrac{u}{u}$$
or, $$\dfrac{u}{v}=\dfrac{u}{f}-1=\dfrac{u-f}{f}$$
or, $$\dfrac{v}{u}=\dfrac{f}{u-f}$$
or, $$m=\dfrac{f}{u-f}$$
or, $$mu=f+mf$$
or, $$f=\dfrac{mu}{1+m}$$