MCQ Questions for Class 10 Physics Light Reflection And Refraction Quiz 5

Magnification produced by a convex mirror is always:

0%
equal to 1
0%
less than 1
0%
more than 1
0%
zero

A ray of light passes from air into water. The angle of refraction will be

0%
Equal to the angle of incidence
0%
Smaller than the angle of incidence
0%
Greater than the angle of incidence
0%
Equal to ${ 45 }^{ \circ }$

Mirror equations for convex mirror and concave mirror are:

0%
Same
0%
Different
0%
Different depending on object position
0%
Same for only a few image positions

The twinkling of stars is due to atmospheric

0%
Refraction of light
0%
Reflection of light
0%
Scattering of light
0%
Dispersion of light

A diverging lens of 20 cm focal length forms on image 15 cm from the lens. What is the distance of the object from the lens?

0%
20 cm
0%
15 cm
0%
60 cm
0%
30 cm

Explanation

Using lens formula as :

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

$\dfrac{1}{15}- \dfrac{1}{20}=\dfrac{1}{u}$

$u= 60 \ cm$

The refractive index of water with respect to air is 1.The refractive index of air with respect to water will be

0%
0.25
0%
0.50
0%
1.00
0%
0.75

An object is placed

$15\ cm$ from a diverging mirror of radius of curvature

$20\ cm$ . What is the image magnification produced?

0%
$+0.4$
0%
$-0.4$
0%
$+2$
0%
$-2$

Explanation

Diverging means convex mirror

Object distance,

$u = -15\ cm$

Image distance,

$v$

Radius of curvature,

$R = 20\ cm$

Focal length,

$f = \dfrac{R}{2} = 10\ cm$

Using mirror formula:

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

$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{15}$

$\Rightarrow \dfrac{1}{v} = \dfrac{1}{6}$

$\Rightarrow v = 6\ cm$

Now, magnification for diverging lens

$m = \dfrac{-v}{u} = - \left (\dfrac{6}{-15} \right )$

${m = 0.4}$
Option A is correct answer

Linear magnification produced by a concave mirror may be

0%
Equal to I
0%
Less than I
0%
More than I
0%
All of the above

Explanation

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.

If linear magnification for a spherical mirror is

$\dfrac{3}{2}$ , then we may write: (symbols have their usual meanings)

0%
$f=\dfrac{u}{2}$
0%
$f=\dfrac{3u}{2}$
0%
$f=\dfrac{3u}{5}$
0%
None of these

Explanation

Mirror equation is:

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

Multiplying both sides by

$u$ , we get:

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

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

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

Now, magnification,

$m=\dfrac{v}{u}=\dfrac{3}{2}$

$\therefore$

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

Solving the above equation we get

$\dfrac{5}{2}f=\dfrac{3}{2}u$

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

The term refraction of light is best described by:

0%
The bending of light rays when they enter from one medium to another medium
0%
Splitting of white light into seven colours when it passes through the prism
0%
Bending of light round corners of obstacles and apertures
0%
Coming back of light from a bright smooth surface

The focal length of a concave mirror depends on:

0%
The radius of curvature of the mirror
0%
The object distance from the mirror
0%
The image distance from the mirror
0%
Both the image and the object distance

Explanation

For any spherical mirror, the focal length

$f$ is equal to half of the radius of curvature

$R$ .

$f=\dfrac{R}{2}$

So, the focal length of a concave mirror depends upon the radius of curvature of the mirror.

The magnification produced by a concave mirror:

0%
is always more than one
0%
is always less than one
0%
is always equal to one
0%
may be less than, equal to or greater than one

Explanation

A concave mirror can form following types of images;

1. Real and diminished when object is kept beyond radius of curvature so, $m$ is negative and $0<|m|<1$ .

2. Virtual and enlarged when the object is kept between focus and the pole so, $m$ is positive and $|m|>1$ .

3. Real and enlarged when the object is kept between focus and radius of curvature so, $m$ is negative and $|m|>1$

4. Real and of same size when the object is kept on the radius of curvature so, $m = -1$ .

When reflection takes place from a spherical mirror, positions of the object and its image are measured from the ......................

0%
poles
0%
focus
0%
radius of curvature
0%
focal plane

Twinkling of stars is on account of

0%
Large distance of stars and storms in air
0%
Small size of stars
0%
Large size of stars
0%
Large distance of stars and fluctuations in the density of air

The radius of curvature of a plane mirror is:

0%
zero
0%
infinite
0%
negative
0%
finite

$\dfrac {\sin i}{\sin r} = constant$ . This is

0%
Newton's law
0%
Snell's law
0%
Both
0%
None

Explanation

$\dfrac {\sin i}{\sin r} = constant$
This constant is known as the

$refractive\ index$ , denoted by n.
This law is known as Snell's law.

A mirage occurs because

0%
The refractive index of atmosphere increases with height
0%
The refractive index of atmosphere decreases with height
0%
The hot ground acts like a mirror
0%
Refractive index remains constant with height

Explanation

$Answer:-$ A

As hot air rises above cold air descends so as we move upward refractive index also decreases .

For exhausted travelers in the desert, an inferior mirage may appear to be a lake of water in the distance. An inferior mirage is called "inferior" because the mirage is located under the real object. The real object in an inferior mirage is the (blue) sky or any distant (therefore bluish) object in that same direction. The mirage causes the observer to see a bright and bluish patch on the ground in the distance.

Light rays coming from a particular distant object all travel through nearly the same air layers and all are bent over about the same amount. Therefore, rays coming from the top of the object will arrive lower than those from the bottom. The image usually is upside down, enhancing the illusion that the sky image seen in the distance is really a water or oil puddle acting as a mirror.

When rays of light fall on a convex lens, it

0%
Converges them
0%
Does not bend them
0%
Diverges them
0%
Enlarges them

If magnification is positive, the nature of the image is:

0%
real and inverted
0%
virtual and erect
0%
real
0%
none of these

Explanation

$\text{Magnification}=\dfrac{\text{Image size}}{\text{Object size}}$

According to new Cartesian sign convention, size of height of real and inverted image is considered negative and that of virtual and erect image is considered positive.

The heigh of the object, being erect is considered positive always.

So, for positive magnification, the ratio, mentioned above, will be positive. This implies that the image height will also be positive.

The image will be virtual and erect.

Twinkling of stars is due to

0%
variation of refractive index in the earth's atmosphere
0%
the fact that light is not emitted by the stars continuously
0%
the absorption oflight by earth's atmosphere
0%
none of these

The principal axis is also called _______ of the lens.

0%
optical axis
0%
x-axis
0%
y-axis
0%
axis

The distance between the extreme points on the periphery of the mirror is called :

0%
focal length
0%
radius of curvature
0%
aperture
0%
pole

If we say that the focal length of a spherical mirror is

$n$ times its radius of curvature, then

$n$ must be

0%
$2.0$
0%
$1.5$
0%
$0.2$
0%
$0.5$

Explanation

According to the question-

$f = nR$

We know the relation between focal length and radius of curvature, which is:

$R = 2f$

S0,

$\Rightarrow f = 0.5 R$

So, value of

$n$ is

$0.5$ .

The ratio of the size of the image to the size of the object is known as :

0%
The Focal Plane
0%
The Transformation ratio
0%
The Efficiency
0%
The Magnification ratio

Explanation

Ratio of the size of the image to the size of object is known as the Magnification ratio. It is given as:

$m = \dfrac{v}{u}$

A convex lens forms a virtual image of an object placed at

$15 cm$ from the lens. The focal length of the lens can be

0%
$2 cm$
0%
$8 cm$
0%
$15 cm$
0%
$18 cm$

When a ray of light passes from an optically less dense medium to a more dense medium, it :

0%
Goes undeviated
0%
Bends towards the normal
0%
Bends away from the normal
0%
Gets reflected

The middle point of a spherical mirror is called

0%
pole
0%
centre of sphere
0%
centre of curvature
0%
none of these

The image of our face in a plane mirror is :

0%
real
0%
magnified
0%
diminished
0%
virtual

The bending of light ray after passing through a medium is commonly known as :

0%
Refraction
0%
Scattering
0%
Reflection
0%
Interference

The distance between the optical centre of lens and the focal point is called :

0%
Reflective index
0%
Focal length
0%
Radius of curvature
0%
Plane of incidence

If a lens diverges a parallel beam of light, the lens is -

0%
Convex
0%
Concave
0%
Cylindrical
0%
Parabolic

Light enters from denser to rarer medium. Choose the correct option.

0%
Angle of refraction = Angle of incidence
0%
Angle of refraction > Angle of incidence
0%
Angle of refraction < Angle of incidence
0%
Angle of refraction $\geq$ Angle of incidence

A concave mirror is made from a hollow sphere of radius of curvature

$30 cm$ . If an object of height

$2 cm$ is placed at

$10 cm$ from the pole of the mirror, determine the size of the image :

0%
$3 cm$
0%
$6 cm$
0%
$12 cm$
0%
$24 cm$

Explanation

Given:

$u=-10cm$

$f=-15cm$

$h_{o}=+2cm$

To find,

$h_{i}$

From mirror formula,

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

$\Rightarrow\dfrac{1}{v}-\dfrac{1}{10}=-\dfrac{1}{15}$

$\Rightarrow \dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{15}=\dfrac{5}{150}$

$v=+30cm$

$magnification=\dfrac{h_{i}}{h_{o}}=\dfrac{-v}{u}$

$\dfrac{h_{i}}{+2}=\dfrac{-30}{-10}$

$h_{i}=+6cm$

If the refractive index of water and glass with respect to air are, respectively,

$\displaystyle \frac {4}{3}$ and

$\displaystyle \frac {3}{2}$ then what will be the refractive index of glass with respect to water?

0%
$\displaystyle \frac {9}{8}$
0%
$\displaystyle \frac {8}{9}$
0%
$\displaystyle \frac {4}{3}$
0%
$\displaystyle \frac {3}{2}$

Explanation

Refractive index of medium 1 w.r.t. medium 2 is given by,

$^{2}\mu_{1} = \dfrac{\mu_{1}}{\mu_{2}}$

Hence according to question,

Refractive index of glass,

$\mu_{g}=\dfrac{3}{2}$

Refractive index of water,

$\mu_{w}=\dfrac{4}{3}$

Hence

$^{w}\mu_{g}=\dfrac{\mu_{g}}{\mu{w}}$

$^{w}\mu_{g}=\dfrac{9}{8}$

An object is placed at 40 cm from the optic centre of a convex lens on its principal axis. if the focal length of the lens is 24 cm, find how far from the lens the screen should be placed to obtain a well-defined image.

0%
48 cm
0%
38 cm
0%
50 cm
0%
60 cm

Explanation

given:

$u=-40cm$

$f=24cm$
solution:from lens formula

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

$\dfrac{1}{v}+\dfrac{1}{40}=\dfrac{1}{24}$

$\dfrac{1}{v}=\dfrac{1}{24}-\dfrac{1}{40}=\dfrac{16}{960}=\dfrac{1}{60}$

$v=60cm$

Calculate the magnification of an object if it is kept at a distance of

$3 cm$ from a concave mirror of focal length

$4 cm$ :

0%
$3$
0%
$6$
0%
$9$
0%
$4$

Explanation

Given:

Object distance

$u=-3cm$

Focal length

$f=-4cm$

By mirror formula,

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

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

$\dfrac{1}{v}=\dfrac{1}{12}$

So, image distance

$v= +12 cm$

Magnification

$m=-\dfrac{v}{u}$

$=-\dfrac{+12}{-3}=+4$

Option D is correct.

When a ray of light strikes a plane mirror along the normal, the angle of incidence is -

0%
$\displaystyle 90^{\circ}$
0%
zero
0%
$\displaystyle 45^{\circ}$
0%
none of these

Explanation

Angle of incidence is the angle made by incident ray and normal to the mirror. Here, the ray incidents along normal so,

From the figure, angle of incidence

$=0^{\circ}$ .

A spherical mirror whose inner hollow surface is the reflecting surface is:

0%
Convex mirror
0%
Concave mirror
0%
Convex lens
0%
Concave lens

The distance between the pole and centre of curvature of a spherical mirror is called _________ .

0%
radius of curvature
0%
focal length
0%
object distance
0%
image height

The angle between reflected ray and the _______ is called angle of reflection

0%
normal
0%
incident ray
0%
reflecting surface
0%
none of the above

Identify the mirror

0%
Convex
0%
Concave
0%
Biconcave
0%
Biconvex

Explanation

$\bf{Convex\ Mirror}$ is a curved mirror where the reflective surface bulges out, towards the light source.

$\bf{Concave\ Mirror}$ is a curved mirror where the reflective surface is curved inwards.

In the figure, the reflective surface is curved inwards. Therefore, it is a concave mirror. Hence, the correct answer is option B.

A spherical mirror with its outer bulging surface as the reflecting surface is a :

0%
convex mirror
0%
concave mirror
0%
plane mirror
0%
none of these

State whether true or false.

The straight line joining the centre of curvature and pole of a spherical mirror is called plane of incidence.

0%
True
0%
False

Fill in the blanks

The ray of light bending away from normal is called _____.

0%
Refracted ray
0%
Incident ray
0%
Emergent ray
0%
None of the above

When light moving in one medium falls at the surface of another medium, _______ light returns back to the same medium.

0%
all of
0%
a part of
0%
no
0%
either all or no

Refraction occurs due to:

0%
Change in direction of light
0%
Change in speed of light
0%
Change in quality of light
0%
All

Explanation

Refraction occurs due to change in optical properties of the light which causes change in direction and speed of light.

The light bends towards the normal or away from the normal if the refracting medium is denser or rarer respectively. The extent of bending is governed by Snell's law:

$\dfrac{n_i}{n_r}=\dfrac{sin \ \theta_r}{sin \ \theta_i}$

The light slows down in a denser medium and becomes faster in a rarer medium.

A man sees a small stationary fish deep in a very clear lake and shoots it accurately. He will

0%
hit the fish
0%
miss the fish
0%
hit but fish will not die
0%
may hit or miss depending on the bullet

Which phenomena of light makes a pool of water appears to be less deep than it actually is?

0%
Reflecion
0%
Refraction
0%
Diffraction
0%
Lateral inversion

Which of the states Snell's law?

0%
Incident ray, refracted ray and the normal at the point of incidence all lies in the same plane
0%
$\dfrac {sin i}{sin r} =$ constant (for a given media)
0%
Angle of incidence $=$ Angle of refraction
0%
All

Explanation

The ratio of sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given colour and for the given pair of media. This law is also known as Snell’s law of refraction.

If i is the angle of incidence and r is the angle of refraction, then,

$\dfrac {\sin i}{\sin r} =$ constant (for a given media)

This constant value is called the refractive index of the second medium with respect to the first.

Fill in the blanks:

A spherical mirror whose reflecting surface is curved outwards is called _________.

0%
concave mirror
0%
convex mirror
0%
plane mirror
0%
biconvex mirror

CBSE Questions for Class 10 Physics Light Reflection And Refraction Quiz 5 - MCQExams.com

0:0:1

Practice Class 10 Physics Quiz Questions and Answers

CBSE Questions for Class 10 Physics Light Reflection And Refraction Quiz 5 - MCQExams.com

0:0:1

Practice Class 10 Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.