MCQ Questions for Class 10 Physics Light Reflection And Refraction Quiz 2

Angle which the normal ray makes with the mirror is equal to:

0%
$\displaystyle { 30 }^{ \circ}$
0%
$\displaystyle { 45 }^{ \circ }$
0%
$\displaystyle { 90 }^{ \circ }$
0%
$\displaystyle { 180 }^{ \circ }$

Explanation

The perpendicular drawn to the surface of the mirror at the point of incidence is known as a normal ray. So, the angle made by the ray with the mirror is equal to

$\displaystyle { 90 }^{ \circ }$ .

The ray which bounces off the surface of the mirror when the incident ray strikes the mirror is known as ________. The point at which incident ray meets the mirror is called the _________.

0%
normal ray, point of reflection
0%
reflected ray, point of refraction
0%
reflected ray, point of incidence
0%
diffracted ray, point of diffraction

When a ray of light passes through a rectangular glass slab, the emergent ray is

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parallel to the incident ray produced
0%
perpendicular to the incident ray produced
0%
oblique to the incident ray produced
0%
inclined to the incident ray produced

Fill in the blanks:

The area of concave mirror that is exposed to incident light is called ________.

0%
aperture
0%
centre of curvature
0%
focus
0%
focal length

Fill in the blanks:

An image that cannot be obtained on a screen is called _________.

0%
real
0%
virtual
0%
plane
0%
none

Concave lenses are also known as

0%
Converging lenses
0%
Diverging lenses
0%
Plane lenses
0%
Circular lenses

Explanation

$Answer:-$ B option

Concave lenses are thinner at the middle. Rays of light that pass through the lens are spread out (they diverge). A concave lens is a diverging lens.

What is known as linear magnification of spherical mirrors?

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Ratio of size of image to size of object
0%
Ratio of shape of image to size of object
0%
Ratio of size of image to shape of object
0%
None

Different types of spherical mirrors are:

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plane and curved
0%
convex and concave
0%
real mirrors
0%
virtual mirrors

Explanation

Spherical mirrors are of

$2$ types:

A spherical mirror, whose reflecting surface is curved inwards is called a concave mirror.
A spherical mirror, whose reflecting surface is curved outwards, is called a convex mirror.

Which of the following is the main reason of Opacity of light waves.

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Reflection
0%
Absorption
0%
Refraction
0%
Diffusion
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Transmission

$\dfrac {\sin i}{\sin r} = ......$

0%
$1$
0%
$2$
0%
Constant
0%
Variable

Explanation

The ratio of the sine of the incident angle

$(i)$ to the sine of the refracted angle

$(r)$ constant for a pair of two media and is equal to the refractive index of the media.

Reflected ray is the ray of light reflected from a surface.

0%
True
0%
False

Normal is an imaginary line perpendicular to the surface where the reflection occurs.

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True
0%
False

A ray of light travels from a denser to a rarer medium then, the ray :

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doesn't bend at all
0%
bends towards the normal
0%
bends away from the normal
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goes along the normal

The height of the image formed by an object of height

$10 cm$ placed in front of a plane mirror is ________.

0%
$5 cm$
0%
$10 cm$
0%
$15 cm$
0%
$20 cm$

Explanation

If an object is placed in front of a plane mirror, the size of the object remains same. This is the property of plane mirror. So the height of the object will be

$10 cm$ only.

When a ray of light travels from air to glass slab and strikes the surface of separation at

$90^o$ , then it ....... .

0%
Bends towards normal
0%
Bends away from normal
0%
Passes unbent
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Passes in zigzag way

The focal length of a spherical mirror is:

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Maximum for red light
0%
Maximum for blue light
0%
Maximum for white light
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Same for all lights

In the case of refraction of light :
a) Frequency changes
b) Speed changes
c) Wavelength changes

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a is correct
0%
b and c are correct
0%
a, b, c are correct
0%
a and b are correct

Explanation

In the process of refraction, speed of light and wavelength of light changes.

In a medium, speed of light is given by

$v=\dfrac{c}{\mu }$ where

$\mu$ is refractive index of the medium. Hence, for medium with different refractive index , speed of light is different.

Frequency will not change because at the boundary/interface of the medium, the number of waves you send is the number of waves you receive at the other side, almost instantly.

Since, for a travelling wave we have a relation

$\nu = \dfrac{v}{\lambda}$ .

Hence wavelength also changes in the process of refraction.

So, option B is correct.

A square wire of side

$3.0 cm$ is placed

$25 cm$ in front of a concave mirror of focal length

$10 cm$ , with its centre on the axis of the mirror and its plane normal to the axis. The area enclosed by the image of the wire is :

0%
7.5 cm $^{2}$
0%
6.0 cm $^{2}$
0%
4.0 cm $^{2}$
0%
3.0 cm $^{2}$

Explanation

We know that the magnification of a concave mirror is given as

$m=\displaystyle\frac{f}{f-u}$ .
So, we get:

$m=\displaystyle\frac{-10}{-10-(-25)}=-\frac{2}{3}$
Thus the length of the side gets magnified by this factor. The image length is thus

$\displaystyle\frac{2}{3}\times 3=2 cm$
Area of the image is thus

$2\times 2=4 cm^2$

A needle of length

$5\ cm$ , placed

$45 cm$ from a lens, forms an image on a screen placed

$90\ cm$ on the other side of the lens. The type of lens and its focal length is:

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Convex, $30\ cm$
0%
Concave, $30\ cm$
0%
Convex, $60\ cm$
0%
Concave, $60\ cm$

Explanation

Object distance,

$u = -45\ cm$

image distance,

$v = 90\ cm$

focal length,

$f$

Using Lens formula,

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

$\dfrac{1}{f} = \dfrac{1}{90} - \dfrac{1}{\left(-45\right)}$

$\dfrac{1}{f} = \dfrac{1 + 2}{90} = \dfrac{3}{90}$

$f = 30\ cm$

Focal length is positive

$\therefore$ The type of lens is a convex lens

The image of an object placed in front of a concave mirror of focal length 12 cm is formed at a point which is 10 cm more distant from the mirror than the object. The magnification of the image is:

0%
$1.5$
0%
$2$
0%
$2.5$
0%
$3$

Explanation

We have:

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

$f = 12\ cm$ and

$v = u + 10$

$\implies \dfrac {1}{12} = \dfrac {1}{u} + \dfrac {1}{u +10}$

$\implies u = 20\ cm$

$\implies v = 20 +10 = 30\ cm$
We have magnification,

$m = \dfrac {v}{u} = \dfrac{30}{20} = 1.5$

A convex lens of focal length

$15\ cm$ produces a virtual image at a distance of

$30\ cm$ from the lens. The object distance is:

0%
$5\ cm$
0%
$7.5\ cm$
0%
$-10\ cm$
0%
$10\ cm$

Explanation

Given, convex lens

Focal length,

$f=15\ cm$

Object distance,

$v=-30\ cm$ (virtual image, behind the lens)

Object distance,

$u$

Using Lens formula,

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

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

$\dfrac{1}{u}=\dfrac{1}{-30}-\dfrac{1}{15}$

$\dfrac{1}{u}=\dfrac{-2-1}{30}$

$\dfrac{1}{u}=\dfrac{-3}{30}$

$u=-10\ cm$

An object is situated at a distance of 15cm from a convex lens of focal length 30 cm. The position of the image formed by it will be

0%
- 30 cm
0%
30 cm
0%
-10 cm
0%
10 cm

Explanation

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

$\dfrac{1}{v}-\dfrac{1}{-15}=\dfrac{1}{30}$

$\dfrac{1}{v}=\dfrac{1}{30}-\dfrac{1}{15}$

$\dfrac{1}{v}=\dfrac{-1}{30}$

$v= -30cm$

Beams of light are incident through the holes A and B and emerge out of box through the holes C and D respectively as shown in the figure. Which of the following could be inside the box?

0%
A rectangular glass slab
0%
A convex lens
0%
A concave lens
0%
A prism

A plane mirror reflects a beam of light to form a real image. The incident beam is :

0%
parallel
0%
convergent
0%
divergent
0%
any one of the above

If a spherical mirror is immersed in a liquid, its focal length will:

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Increase
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Decrease
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Remain unchanged
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Depend on the nature of liquid

An object is placed at a distance

$2 f$ from the pole of a convex mirror of focal length

$f$ . The linear magnification is:

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$\dfrac{1}{3}$
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$\dfrac{2}{3}$
0%
$\dfrac{3}{4}$
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$1$

Explanation

Here,

Object distance, $u=-2f$

Image distance, $v=$ ?

From mirror formula,

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

$\dfrac{1}{v}+\dfrac{1}{-2f}=\dfrac{1}{f}$

$\dfrac{1}{v}=\dfrac{3}{2f}$

$v=\dfrac{2f}{3}$

Now,
Magnification, $M=\dfrac{-v}{u}$

$=-\dfrac{\dfrac{2f}{3}}{(-2f)}$

$=\dfrac{1}{3}$

(M is positive for convex mirror)

The angle of reflection is given by the angle :

0%
NOQ
0%
QOB
0%
PON
0%
AOP

The line PO is called the _______.

0%
Reflected ray
0%
Incident ray
0%
Normal
0%
Refracted ray

The line OQ is called the :

0%
reflected ray
0%
incident ray
0%
normal
0%
refracted ray

A light ray is traveling from air to glass. The angle of incidence on the boundary is 30

$^{\circ}$ . Find the sine of the angle of refraction. (Take:

$sin\ 30^o = \dfrac{1}{2}$ )

0%
1/3
0%
2
0%
3
0%
1

Explanation

Given:

The angle of incidence is

$i=30^o$

Refractive index of glass is

$n_g=1.5$

Refractive index of air is

$n_a=1$

The sine angle of refraction

$sin\ r=?$

using Snell's law we can write,

$n_a\times sin\ i = n_g \times sin\ r$

$\therefore sin\ r = \dfrac{n_a\times sin\ i}{n_g}=\dfrac{1\times sin\ 3o^o}{1.5}= \dfrac{1}{3}$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point 15 cm from the lens on the opposite side. If the lens is removed the point where the rays meet will move 5 cm closer to the lens. The focal length of the lens is :

0%
5 cm
0%
-10 cm
0%
20 cm
0%
-30 cm

Explanation

From lens formula,

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

$\dfrac{1}{15}-\dfrac{1}{10}=\dfrac{1}{f}$

Focal length,

$f=-30cm$

An object

$5$ cm tall is placed

$1$ m from a concave spherical mirror which has a radius of curvature of

$20$ cm. The size of the image is:

0%
$0.11 cm$
0%
$0.50 cm$
0%
$0.55 cm$
0%
$0.60 cm$

Explanation

Here,

Object distance,

$u=-1\ m$

Focal length,

$f=-10\ cm$

Using mirror formula,

$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$ , we get:

$\dfrac{1}{-100}+\dfrac{1}{v}=\dfrac{1}{-20}$

or,

$\dfrac{1}{v}=\dfrac{-9}{100}$

or,

$v=\dfrac{-100}{9}$

Magnification,

$M=\dfrac{-v}{u}=\dfrac{h_{2}}{h_{1}}$

or,

$-\dfrac{\dfrac{100}{9}}{100}=\dfrac{h_{2}}{h_{1}}$

or,

$h_{2}=\dfrac{-5}{9}=-0.55\ cm$

It is desired to photograph the image of an object placed at a distance of

$3$ m from a plane mirror. The camera, which is at a distance of

$4.5$ m from the mirror should be focused for a distance of :

0%
$3 m$
0%
$4.5 m$
0%
$6 m$
0%
$7.5 m$

Explanation

The distance between the object and the mirror is

$3 m$ , i.e.,

$d_1 = 3 m$ ,

Hence image distance also equals

$d_1 = 3 m$

Also, the distance between the mirror and the camera is

$4.5 m$ , i.e.,

$d_2 = 4.5 m$ .

Hence the focus of the camera,

$f = d_1 + d_2 = 3 + 4.5 = 7.5 m$

Statement- I : The formula connecting

$u$ ,

$v$ and

$f$ for a spherical mirror is valid only for mirrors whose sizes are very small compared to their radii of curvature.
Statement- II : Laws of reflection are strictly valid for plane surfaces, but not for large spherical surfaces

0%
Statement -I is true, Statement -II is true; Statement -II is a correct explanation for statement -I
0%
Statement-I is true, Statement -II is true; Statement-II is not a correct explanation for statement- I
0%
Statement - I is true, Statement -II is false
0%
Statement - I is false, Statement -II is true

The line NO that is perpendicular to the plane mirror surface and drawn at the point of incidence is called the :

0%
reflected ray
0%
incident ray
0%
normal
0%
refracted ray

The point C is the :

0%
Center of curvature
0%
Principal focus
0%
Pole
0%
None

Which of the following defines the center of curvature of a curved mirror?

0%
The center of the hollow glass sphere of which the curved mirror was (previously) a part
0%
The geometric center of the curved mirror
0%
The radius of the hollow glass sphere of which the curved mirror was (previously) a part
0%
None of these

The mirror shown is _______ in shape.

0%
Convex
0%
Concave
0%
Concavo-convex
0%
Convexo-concave

The given figure shows a ray of light as it travels from medium A to medium B. Refractive index of the medium B relative to medium A is:

0%
$\dfrac{\sqrt{3}} {\sqrt{2}}$
0%
$\dfrac{\sqrt{2}} {\sqrt{3}}$
0%
$\dfrac{1} {\sqrt{2}}$
0%
$\sqrt{2}$

Explanation

Answer is A.

The Snell's law simply relates angles

$i$ and

$r$ to the refraction indices of the two media A and B. Therefore, refractive index of medium B relative to medium A is:

$n=\dfrac { sin\quad i }{ sin\quad r }=\dfrac{sin\ 60^0}{sin\ 45^0} =\dfrac { \sqrt { 3 } }{ \sqrt { 2 } }$ .

The imaginary line passing through the pole and the center of curvature of the curved mirror is called its _______.

0%
Focal length
0%
Radius of curvature
0%
Principal axis
0%
None

The point F is the :

0%
Principal axis
0%
Principal focus
0%
Pole
0%
None

A convex mirror has a focal length f. A real object is placed at a distance f/2 in front of it from the pole. The mirror produces an image at

0%
$3f$
0%
$f$
0%
$2f$
0%
$f/3$

Explanation

Given,

Focal length

$=f$

Object distance

$u=-\dfrac f2$

Lens formula,

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

$\displaystyle \cfrac{1}{v} - \cfrac{2}{f} =\dfrac{1}{f}$

$\cfrac{1}{v} =\cfrac{-3}{f}$

$v=\cfrac{f}{3}$

Option D

A beam of light is incident through the holes on side A and emerges out of the holes on the other face of the box as shown in the fogure. Which of the following could be inside the box ?

0%
Concave lens
0%
Rectangular glass slab
0%
Prism
0%
Convex lens

Which is the path of rays of light when it enters with oblique incidence from water into air ?

The path of a ray of light passing through a rectangular glass slab tested by four students are shown in the figure as

$A, B, C,$ and

$D$ . Which one of them is correct?

0%
$A$
0%
$B$
0%
$C$
0%
$D$

A student focussed the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle screen and the lens as under :

Position of candle

$= 12.0\ cm$
Position of convex lens

$= 50.0\ cm$
Position of the screen

$= 88.0\ cm$

What will be the nature of the image formed if he further shifts the candle towards the lens?

0%
Virtual and erect
0%
Real and inverted
0%
Real and diminished
0%
None of the above

Explanation

Before shifting the candle,

$u=12-50=-38cm$

$v=88-50=38cm$

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

$\cfrac{1}{f}=\dfrac{1}{38}+\dfrac{1}{38}$

$\cfrac{1}{f}=\dfrac{1+1}{38}$

$f=19\ cm$

Hence, before shifting the candle, it is placed at a distance of

$2f$ from the mirror.

When candle is shifted towards the lens from this position, its distance from the lens is

$f<u<2f$ . In this scenario, image formed is virtual, diminished and erect.

A small hole P is made in a piece of cardboard. The hole is illuminated by a torch as shown in Fig. 16.The pencil of light coming out of the hole falls on a mirror.
At which point should the eye be placed so that the hole can be seen?

0%
A
0%
B
0%
C
0%
D

Explanation

According to law of reflection we know in case of plane mirror the angle of reflection is equal to the angle of incidence. So,if we place the eye on point

$A$ the hole can be seen.

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

In case of a real and inverted image, the magnification of a mirror is:

0%
Positive
0%
Negative
0%
Zero
0%
Infinity

Explanation

We know,
Magnification(M)

$=\dfrac{height \ of\ image({h}_{i})}{height\ of \ object({h}_{o})}$

Here, image is inverted so according to sign convention

$height \ of \ image({h}_{i})$ will be negative while object height will be positive.
Hence, the magnification of a mirror is negative.

CBSE Questions for Class 10 Physics Light Reflection And Refraction Quiz 2 - MCQExams.com

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Practice Class 10 Physics Quiz Questions and Answers

CBSE Questions for Class 10 Physics Light Reflection And Refraction Quiz 2 - MCQExams.com

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Practice Class 10 Physics Quiz Questions and Answers

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