MCQ Questions for Class 10 Physics Light Reflection And Refraction Quiz 11

In a car a rear view mirror having a radius of curvature 1.50 m forms a virtual image of a bus located 10.0 m from the mirror. The factor by which the mirror magnifies the size of the bus is close to

0%
0.06
0%
0.07
0%
0.08
0%
0.09

Explanation

The rear view mirror is a convex mirror.

Given:

Radius of curvature

$R=+1.50 \ m$

Focal length

$f=\dfrac{R}{2}=\dfrac{+1.50}{2}=+0.75 \ m$

Object distance

$u=-10 \ m$

By mirror formula,

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

$\Rightarrow\dfrac{1}{0.75}=\dfrac{1}{v}+\dfrac{1}{-10}$

$\Rightarrow \dfrac{1}{v}=\dfrac{10+0.75}{7.5}=\dfrac{10.75}{7.5}$

$\Rightarrow v=\dfrac{750}{1075}=\dfrac{30}{43} \ m$

Magnification

$m=-\dfrac{v}{u}=-\dfrac{30}{43} \times \dfrac{1}{(-10)}$

$\therefore m=+0.069 \approx +0.07$

Option B is correct.

From the understanding of cartesian sign convention for reflection by spherical mirror, students took part in group discussion for

$FA-I$ in classroom. Who is wrong in the group discussion?
Alpesh : All the distances are measured from the pole of a mirror parallel to the principal axis.
Beena: The distance measured in the direction of incident ray are taken positive.
Champak: The height measured upward and perpendicular to principal axis is taken negative.
Daksha : The height measured downward and perpendicular to principal axis is taken as positive.

0%
Champak and Daksha
0%
Only Champak
0%
Alpesh and Beena
0%
Only Daksha

The magnification produced by a mirror is

$+\dfrac{1}{3}.$ Then the mirror is a ____________.

0%
Concave mirror
0%
Convex mirror
0%
Plane mirror
0%
plano convex mirror

Explanation

The value of

$m$ is less than

$1$ . So it means the image is diminished.

$m=+\dfrac{1}{3}$

The positive sign indicates that the image is virtual and erect.

Hence, the mirror must be

$convex$ mirror. Concave mirror also produces virtual image but it is enlarged. But convex mirror always produces diminished, virtual and erect image.

Answer-(B)

Choose the correct answer from the alternatives given.
There is certain material developed in laboratories which have a negative refractive index. A ray incident from the air (medium 1) into such a medium (medium 2) shall follow a path given by:

Explanation

The negative refractive index materials are those in which incident rays from the air (medium 1) refracts or bends differently - opposite and symmetric to normal to that of positive refractive index medium.

Using Snell's law

$\mu = \dfrac{sin \, i}{sin \, r} \Rightarrow sin \, r = \dfrac{sin \, i}{\mu}$
According to the question,

$\mu$ is negative

$\therefore sin \, r$ is negative.
Hence, r is negative

So light ray will bend as shown in figure

So the answer is (A)

1976689_943925_ans_062aa8f2a9cd4023830a134bd0c93a4f.png

The image of the needle placed 45 cm from a lens is formed on a screen placed 90 cm on the other side of the lens. The displacement of the image, if the needle is moved by 5.0 cm away from the lens is

0%
10 cm, toward the lens
0%
15 cm, away from the lens
0%
15 cm, towards the lens
0%
10 cm, away from the lens

Explanation

Here, u = -45 cm, v = 90 cm

$\therefore$

$\dfrac{1}{f} = \dfrac{1}{v}-\dfrac{1}{u} = \dfrac{1}{90} + \dfrac{1}{45} =\dfrac{1}{30} \Rightarrow f = 30 cm$
when the needle is move 5 cm away from the lens,
u=-(45 + 5) = -50 cm

$\therefore$

$\frac{1}{v'} = \dfrac{1}{f} +\dfrac {1}{u'} = \dfrac{1}{30} + \dfrac{1}{(-50)} = \dfrac{2}{150} = \Rightarrow v' = 75 cm$

$\therefore$ Displacement of image = v-v' =90-75= 15 cm, towards the lens

A convex mirror forms an image one-fourth the size of the object. If object is at a distance of

$0.5\ m$ from mirror the focal length of the mirror is

0%
$0.16\ m$
0%
$-1.5\ m$
0%
$0.4\ m$
0%
$-0.4\ m$

Explanation

Given :

$u = -0.5 \ m$
Convex mirror forms erect image of real object.
Using,

$\dfrac{h_i}{h_o} = -\dfrac{v}{u}$
Or

$\dfrac{1}{4} = -\dfrac{v}{(-0.5)}$

$\implies \ v = 0.125 \ m$
Using mirror formula,

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

$\therefore$

$\dfrac{1}{0.125}+\dfrac{1}{-0.5} = \dfrac{1}{f}$

$\implies \ f = 0.16 \ m$

Refraction of the light from the air to glass and from air to water are shown in the figure (i) and figure (ii) below. The value of an angle

$\theta$ in the case of refraction as shown in the figure (iii) will be

0%
$30$
0%
$35$
0%
$60$
0%
$41$

Explanation

From snell's law,

From air to glass,

(i)

$\mu_{glass}=\cfrac{\sin 60^{\circ}}{\sin 35^{\circ}}$

From Air to water,

$\mu_{water}=\cfrac{\sin 60^{\circ}}{\sin 41^{\circ}}$

Now, From water to glass

On taking ratio of above equation,

$\cfrac{\mu_{glass}}{\mu_{water}}=\cfrac{\sin 60^{\circ}}{\sin 35^{\circ}}\cdot \cfrac{\sin 41^{\circ}}{\sin 60^{\circ}}=\cfrac{\sin 41^{\circ}}{\sin 35^{\circ}}$

Also

$\cfrac{\mu_{glass}}{\mu_{water}}=\cfrac{\sin 41^{\circ}}{\sin 35^{\circ}}=\cfrac{\sin i}{\sin r}$

so, where

$i=41^{\circ}, r=35^{\circ}$

1153102_952203_ans_41166797aa504692a6ac87a04710900e.png

Magnification produced is +

$\dfrac { 1 }{ 3 }$ , then what kind of mirror it is?

0%
concave mirror
0%
convex mirror
0%
opaque mirror
0%
plane mirror

A ray

$R_{1}$ is incident on the plane surface of the glass slab (kept in air) of refractive index

$\sqrt {2}$ at an angle of incidence equal to the critical angle for this air glass system. The refracted ray

$R_{2}$ undergoes partial reflection and refraction at the other surface. The angle between reflected ray

$R_{3}$ and the refracted ray

$R_{4}$ at that surface is :

0%
$45^{\circ}$
0%
$135^{\circ}$
0%
$105^{\circ}$
0%
$75^{\circ}$

Explanation

Since, it is given that angle of incidence is equal to the critical angle

Therefore,

$\sin \,i\,=\,\dfrac{1}{n}\,\,=\,\dfrac{1}{\sqrt{2}}$

$i\,=\,\,45{}^\circ$

According to laws of refraction

$\dfrac{\sin i}{\sin r}\,=\,n\,\,(refractive\,index)$

$sin\,r\,=\,\dfrac{\sin \,i}{n}\,=\,\,\frac{1}{2}$

$r\,=\,30{}^\circ$

Let $\theta$ is the angle between R₃and R₄

So, $\theta \,=\,180{}^\circ \,-\,(i\,+\,r)\,$

$\,\,\,\,\,=\,180\,-\,(\,45\,+\,30\,)$

$\,\,\,\,\,\,=\,\,105{}^\circ$

An object is placed at a distance of 1.5 m from a screen and a convex lens is interposed between them. The magnification produced isThe focal length of the lens is then

0%
1 m
0%
0.5 m
0%
0.24 m
0%
2 m

Explanation

Here,

$m =\dfrac{v}{u} =-4$

$\Rightarrow u = \dfrac{-v}{4}$ ------------(1)

Also,

$|u| + |v| = 1.5$

$\dfrac{v}{4} + v = 1.5$

$\Rightarrow \dfrac{(v+4 v)}{4} = 1.5$

$\Rightarrow v = 1.2 m$

So, putting the value of

$v$ in equation (1):

$u= \dfrac{-1.2}{4} = -0.3 m\,\,\,$

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

$\Rightarrow f=\dfrac{(-0.3 \times 1.2)}{(-0.3 - 1.2)} = 0.24m$
Hence the correct option is

$(C)$

Refraction of light from air to glass and from air to water are shown in figure (i) and figure (ii) below. The value of the angle

$\theta$ in the case of refraction as shown in figure (iii) will be

0%
$30^o$
0%
$35^o$
0%
$60^o$
0%
$41^o$

Explanation

Using Snell's law,

$^au_g \, = \, \dfrac{sin \, 60^{\circ}}{sin \, 35^{\circ}}, \, ^a\mu_w \, = \, \dfrac{sin \, 60^{\circ}}{sin \, 41^{\circ}} \, and \, ^w\mu_g \, = \, \dfrac{sin \, 41^{\circ}}{sin \, \theta}$

$^a\mu_w \, \times \, ^w\mu_g \, = \, ^q\mu_g$

$\Rightarrow \, \dfrac{sin \, 60^{\circ}}{sin \, 41^{\circ}} \, \times \, \dfrac{sin \, 41^{\circ}}{sin \, \theta} \, = \, \dfrac{sin \, 60^{\circ}}{sin \, 35^{\circ}} \, \Rightarrow \, sin \theta \, = \, sin35^{\circ} \, or \, \theta \,= \, 35^{\circ}$

A convex lens of focal length

$f$ produces a real image

$\mu$ times the size of an object, Then what is the distance of the object from the lens?

0%
$\left( \mu -1 \right) f$
0%
$\left( \mu +1 \right) f$
0%
$\cfrac { \left( \mu -1 \right) f }{ \mu }$
0%
$\cfrac { \left( \mu +1 \right) f }{ \mu }$

Explanation

Since image formed is real , so magnification will be negative.

Magnification

$m=-\mu$

$,-\mu =\dfrac{v}{u}$

$,\mu =\dfrac{-v}{u}$

$\Rightarrow v=-u\left( \mu\right)$

Lens formula,

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

$\Rightarrow \dfrac{-1}{u\left( \mu\right) }-\dfrac{1}{u}=\dfrac{1}{f}$

$\Rightarrow u=\dfrac{f\left( \mu +1\right)}{\mu}$

Hence, the answer is

$\dfrac{f\left( \mu +1\right)}{\mu}.$

The distance between an object and its doubly magnified image by a concave mirror is: [ Assume

$f$ = focal length]

0%
$3 f/2$
0%
$2 f/3$
0%
$3f$
0%
Depends on whether the image is real or virtual.

Explanation

The magnification is given as,

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

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

$v = - 2u$

Ignoring the sign and using mirror formula, we get

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

$\dfrac{1}{{2u}} + \dfrac{1}{u} = \dfrac{1}{f}$

$\dfrac{{1 + 2}}{{2u}} = \dfrac{1}{f}$

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

Here, difference between object distance and image distance is also $u$ .

For an isosceles prism of angle

$A$ and refractive index

$\mu$ , it is found that the angle of minimum deviation

$\delta _{m}=A$ . Which of the following option is/are correct?

0%
At minimum deviation, the incident angle $r_{1}$ and the refracting surface are related by $r_{1}=(i_{1}/2)$ .
0%
For this prism the refractive index $\mu$ at the angle of prism $A$ are related as $A=\dfrac {1}{2} \cos^{-1}(\mu/2)$ .
0%
For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $i_{ 1 }=\sin ^{ -1 } \left[ \sin { A } \sqrt { 4\cos ^{ 2 }{ \dfrac { A }{ 2 } } -1-\cos { A } } \right]$ .
0%
for the angle of incidence $I_{1}=A$ ,the ray inside the prism is parallel to the base of the prism.

An object of length

$6\ cm$ is placed on the principle axis of a concave mirror of focal length

$f$ at a distance of

$4\ f$ . The length of the image will be

0%
- $2\ cm$
0%
$12\ cm$
0%
$4\ cm$
0%
$1.2\ cm$

Explanation

Given that,

The object distance $u=-6\,cm$

Now, magnification is

$m=\dfrac{I}{O}$

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

$\dfrac{I}{6}=\dfrac{-f}{-f-\left( -4f \right)}$

$I=-2\,cm$

Hence, the length of image is - $2\ cm$

A bulb is located on a wall. Its image is to be obtained on a parallel wall with the help of convex lens. If the distance between parallel walls is

$D$ then required focal length of lens placed in middle between the walls is :

0%
only $\dfrac { D}{ 4 }$
0%
only $\dfrac { D }{ 2 }$
0%
More than $\dfrac { D }{ 4 }$ but less than $\dfrac { D }{ 2 }$
0%
less than or equal to $\dfrac { D }{ 4 }$

Explanation

Refer above image,

$u=-D/2$

$v=D/2$

$f=?$

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

$\dfrac{1}{f}=\dfrac{1}{D/2}-\dfrac{1}{-D/2}$

$\dfrac{1}{f}=2/D+2/D$

$\dfrac{1}{f}=\dfrac{4}{D}$

$f=\dfrac{D}{4}$

2081146_1110705_ans_0eed5c87ce794c319a6e29a272ea4f20.png

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 :

0%
$-5.5\ cm$
0%
$-6.5\ cm$
0%
$- 7.5\ cm$
0%
$- 8.5\ cm$

Explanation

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).

1358538_1133135_ans_7fa0e44964964bd0848c687b68a5d327.png

The magnification of plane mirror is always -

0%
$<1$
0%
$> 1$
0%
$= 1$
0%
Zero

Explanation

The size of the image is equal to size of the object for plane mirror.

So, magnification is equal to ratio of image height to object height so it will subsequently be $1$ .

An infinitely long rod lies along the axis of a concave mirror of focal length l. The near end of the rod is at distance u

$>$ f from the mirror. Its image will have a length?

0%
$\dfrac{uf}{u-f}$
0%
$\dfrac{uf}{u+f}$
0%
$\dfrac{f^1}{u-f}$
0%
$\dfrac{f^2}{u-f}$

Explanation

The mirror formula is given by the equation

$\dfrac{1}{f}$ =

$\dfrac{1}{v}$ +

$\dfrac{1}{u}$ ---------(1)

where f=focal length, v=image distance and u =object distance

As the condition is for mirror its object distance and focal length both lies on left side of mirror thus we consider it as negative where as its image forms behind the mirror we consider it positive.

Hence equation 1 can be reduced as,

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

$v=\dfrac{uf}{f-u}$ -------(2)

Now length of the image

$L=|v|-|f|$

$\dfrac { uf }{ f-u } -f\\ =\dfrac { uf-{ f }^{ 2 }+uf }{ f-u } \\ =\dfrac { { -f }^{ 2 } }{ f-u }$

$L=\dfrac { { f }^{ 2 } }{ u-f }$

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 ?

0%
$3\ cm$
0%
$2.4\ cm$
0%
$10\ cm$
0%
$5\ cm$

Explanation

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.

1383148_1211616_ans_df6df4b67eb04cbb8cb41b7539c6dea4.png

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 ?

0%
$30.7^{o}$
0%
$35.3^{o}$
0%
$41.7^{o}$
0%
$48.6^{o}$

For a mirror linear magnification m come out to be +What conclusions can be drawn from this ?

0%
Mirror is concave
0%
Mirror can be convex or concave but it cannot be plane
0%
Object lies between pole and focus
0%
Object lies beyond focus

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

0%
Plane mirror
0%
Concave mirror
0%
convex mirror
0%
None of these.

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-

0%
$1600 sq. cm$
0%
$800 sq. cm$
0%
$400 sq. cm$
0%
$200 sq. cm$

Explanation

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

$=16\times 100=1600$

${ cm }^{ 2 }$ .

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

0%
$30\ cm$
0%
$90\ cm$
0%
$120\ cm$
0%
$60\ cm$

Explanation

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

$\left| V \right| = \dfrac{{\left| u \right|}}{u}$

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

$.$

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

0%
$75 \ cm$
0%
$60 \ cm$
0%
$125 \ cm$
0%
$50 \ cm$

Explanation

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.

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

0%
Virtual, erect behind the mirror at 8.9 cm
0%
Virtual, erect behind the mirror at 89 cm
0%
Real, erect behind the mirror at 8.9 cm
0%
Virtual, inverted behind the mirror at 89 cm

Explanation

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.

A real object is placed at

$20 \ cm$ from a concave mirror of focal length

$5 \ cm$ . Find the magnification.

0%
$3$
0%
$-3$
0%
$\cfrac{1}{3}$
0%
$-\cfrac{1}{3}$

Explanation

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.

2072681_1312124_ans_046c8d8821ce44e2ba52c2b2639535ad.png

A ray of light is incident on the surface of separating two transparent medium at an angle

$45^0$ and is refracted in medium at an angle

$30^0$ . Velocity of light in the medium will be...

0%
$3.8 \times 10^8 \ m/s$
0%
$3.38 \times 10^8 \ m/s$
0%
$2.12 \times 10^8 \ m/s$
0%
$1.5 \times 10^8 \ m/s$

A glass containing a liquid appears to be half filled when viewed from top, if it is actually two-thirds filled. The refractive index of liquid is

0%
$\cfrac 7 6$
0%
$\cfrac 5 4$
0%
$\cfrac 4 3$
0%
$2$

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.

0%
$2 \mu s$
0%
$1.5 \mu s$
0%
$1 \mu s$
0%
$3 \mu s$

A ray of light is travelling through a medium of refractive index

$\sqrt { 2 }$ with respect to air. When it is incident on the surface making an angle

${ 45 }^{ o }$ with the surface, which of the following will take place?(Medium to which refracted ray travels is air)

0%
angle of refraction will be ${ 45 }^{ o }$ .
0%
angle of refraction will be ${ 90 }^{ o }$
0%
The ray will be reflected internally
0%
None of the above

A convex lens is placed somewhere in between an object and a screen. The distance between the object and screen is

$48\ cm$ . If the numerical value of the magnification produced by the lens is

$3$ , the focal length of the lens is :

0%
$16\ cm$
0%
$4.5\ cm$
0%
$12\ cm$
0%
$9\ cm$

Explanation

The distance between object and screen

$= 48\ cm$

$u + v = 48\ cm$ ...............(i)

Magnification

$= 3$

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

$\boxed{v = 3u}$

Putting the value in equation (i)

$u + 3u = 48$

$\boxed{u = 12\ cm}$

$\boxed{v = 3 \times 12 = 36 cm}$

Focal length is given by

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

$= \dfrac{1}{36} - \dfrac{1}{-12}$

$\dfrac{1}{f} = \dfrac{1 + 3}{36}$

$\boxed{f = 9\ cm}$

2096703_1446368_ans_244f9011cbd94daeb7e1923cfb6fa901.png

A ray of light is incident on a glass at an angle of

$60^o$ . What would be the refractive index of glass, if reflected and refracted rays are perpendicular to each other ?

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

Explanation

The incident ray $1$ will be reflected at an equal angle as reflected ray $2.$

By laws of reflection, $i=r=60^o$

The refracted ray $3$ is perpendicular to reflected ray $2.$

So by geometry, the angle of refraction $r'=30^o$

Now by Snell's law,

$\mu_{air} \ sin \ i= \mu_{glass} \ sin \ r'$

$(1) \times sin \ 60^o= \mu_{glass} \times sin \ 30^o$

$\dfrac{\sqrt3}{2}=\mu_{glass} \times \dfrac{1}{2}$

$\mu_{glass}=\sqrt3$

Option A is the answer.

1996023_1441036_ans_e852a520e2664e748aa585b02a1735a0.JPG

The refractive indices of glass and water are 3/ 2 and 4/ 3 respectively. The refractive index of glass with respect to water is

0%
2
0%
8/9
0%
9/8
0%
5/3

Explanation

Given

$n_{glass}=3/2$

$n_{water}=4/3$

Now refractive index of glass with respect to water is given by

$_{water}n_{glass}$ =

$\dfrac{n_{glass}}{n_{water}}$ =

$\dfrac{3/2}{4/3}$ =

$\dfrac{3\times 3}{4 \times 2}$ =

$\dfrac{9}{8}$

Monochromatic light is refracted from air into the glass of refractive index

$\mu$ . The ratio of the wavelength of incident and refracted waves is

0%
1: $\mu$
0%
1: $\mu$ $^{2}$
0%
$\mu$ :1
0%
1:1

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-

0%
$30cm$
0%
$90 cm$
0%
$120 cm$
0%
$60 cm$

Explanation

Focal length of the convex mirror (f) = 30 cm

${h_i} = \frac{{{h_0}}}{4}$

$\Rightarrow m=\frac{{{h_i}}}{{{h_0}}} = \frac{1}{4} = - \frac{v}{u}=m$

$\Rightarrow v = - \frac{u}{4}$

The mirror formula is given as

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

Substituting the values in the formula,

$\frac{1}{{30}} = \frac{1}{u} - \frac{4}{u}$

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

$u = - 90{\text{ cm}}$

Therefore, the distance of the object from the mirror is 90 cm.

The ray of light emerging out of glass slab is always parallel to the incident ray.

0%
True
0%
False

${ n }_{ 12 }\times{ n }_{ 21 },$ in simplest from is

0%
1
0%
2
0%
0
0%
4

Explanation

Suppose ${\eta _1}$ and ${\eta _2}$ is the refractive index of medium $1$ and $2$ .

The refractive index of the medium $1$ with respect to the medium $2$ is given as

${\eta _{12}} = \dfrac{{{\eta _1}}}{{{\eta _2}}}$

Similarly the refractive index of the medium $2$ with respect to the medium $1$ is given as

${\eta _{21}} = \dfrac{{{\eta _2}}}{{{\eta _1}}}$

Hence,

${\eta _{12}} \times {\eta _{21}} = \dfrac{{{\eta _1}}}{{{\eta _2}}} \times \dfrac{{{\eta _2}}}{{{\eta _1}}}$

${\eta _{12}} \times {\eta _{21}} = 1$ .

Two objects A and B when placed in placed in turn in front of a concave mirror of focal length 7.5 cm give images of equal size. If A is three times the size of B and is placed 30 cm from the mirror, find the distance of B from the mirror.

0%
-15 cm
0%
25 cm
0%
-17.5 cm
0%
-11 cm

Explanation

Given,

The size of the image of object A and B is the same, hence

${h_A}^\prime = {h_B}^\prime$

The size of object A is four times that of B, hence

${h_A} = 3{h_B}$

Now the magnification of object A can be given as

${m_A} = \dfrac{{{h_A}^\prime }}{{{h_A}}}$ …… (1)

Similarly for object B

${m_B} = {\text{ }}\dfrac{{{h_B}^\prime }}{{{h_B}}}$ ……. (2)

From equation (1) and (2) the ratio of magnification is given as

$\dfrac{{{m_A}}}{{{m_B}}} = \dfrac{{{h_A}^\prime }}{{{h_A}}}{\text{ }} \times {\text{ }}\dfrac{{{h_B}}}{{{h_B}^\prime }} \Rightarrow \dfrac{{{m_A}}}{{{m_B}}} = \dfrac{{{h_A}^\prime }}{{3{h_B}}}{\text{ }} \times {\text{ }}\dfrac{{{h_B}}}{{{h_A}^\prime }}$

$\therefore \dfrac{{{m_A}}}{{{m_B}}} = \dfrac{1}{3}$ ……. (3)

Now the magnification formula is given as

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

Therefore equation (3) becomes

$\dfrac{{{m_A}}}{{{m_B}}} = \dfrac{{f - {u_B}}}{{f - {u_A}}} \Rightarrow \dfrac{1}{3} = \dfrac{{ - 7.5 - {u_B}}}{{ - 7.5 + 30}}$

$- {u_B} = \dfrac{{22.5}}{3} + 10 \Rightarrow {u_B} = - 17.5cm$

Hence the object B is placed at $-17.5\;cm$ from the mirror.

Refractive index of a rectangular glass slab is

$\mu = \sqrt { 3 }.$ A light ray incident at an angle

$60 ^ { \circ }$ is displaced laterally through

$2.5 { cm } .$ Distance travelled by light in the slab is

0%
$4 cm$
0%
$5 cm$
0%
$2.5\sqrt { 3 } { cm }$
0%
$3 cm$

Explanation

Given, R.I. of the glass slab

$\mu=\sqrt3$ ,

Angle of incidence at the first surface of the glass,

$\theta_i=60^\circ$ .

We know, lateral shift,

$d=t\dfrac{sin(\theta_i-\theta_e)}{cos\theta_e}=2.5cm$ (given).

Again, from geometry,

$t=lcos\theta_e$ .

Here,

$l$ be the distance travelled by light in the slab,

And,

$\theta_e$ and

$t$ be the angle of refraction at the first surface and depth of the slab.

So, we have,

$d=lsin(\theta_i-\theta_e)$ ......... (1)

Now, using Snell's law,

$sin\theta_i=\mu sin\theta_e$

Or,

$sin60^\circ=\sqrt3 sin\theta_e$

Or,

$sin\theta_e=\dfrac{1}{2}$

Or,

$\theta_e=30^\circ$

Now, from the equation (1),

$2.5=l\times sin(60^\circ -30^\circ)=l\times sin30^\circ$

Or,

$l=5cm$

So, the correct option is (B).

A thin glass (refractive index 1.5) lens has optical power of -5D in air.Its optical power in a liquid medium with refractive index 1.6 will be:

0%
-1D
0%
1 D
0%
25 D
0%
-25 D

The refractive index of material of a prism of angles

$45^{\circ}, -45^{\circ}$ , and

$-90^{\circ}$ is

$1.5$ . The path of the ray of light incident normally on the hypotenuse side is shown in

Sunlight that bounces off a surface is said to be __________ from the surface.

0%
Radiated
0%
Absorbed
0%
Emitted
0%
Reflected

If the light moving in a straight line bends by a small but fixed angle, it may be a case of

0%
reflection
0%
refraction
0%
diffraction
0%
dispersion

Light propagates

$2\ cm$ distance in glass of refractive index

$1.5$ in time

$t_0$ . In the same time

$t_0$ , light propagates a distance of

$2.25\ cm$ in a medium. The refractive index of the medium is :

0%
$4/3$
0%
$3/2$
0%
$8/3$
0%
$none\ of\ these$

Explanation

In medium-1 (refractive index

$n_1=1.5$ ), light travels a distance

$2\ cm$ in time

$t_0$ .

The speed of light in medium-1 is

$v_1 = \dfrac{2\times 10^{-2}\ m}{t_0\ sec}$

In medium-2 (refractive index

$n_2$ ), light travels a distance

$2.25\ cm$ in time

$t_0$ .

The speed of light in medium-2 is

$v_2 = \dfrac{2.25\times 10^{-2}\ m}{t_0\ sec}$

From the definition of refractive index, we can write,

$\dfrac{n_1}{n_2}=\dfrac{v_2}{v_1}$

$\Rightarrow \dfrac{1.5}{n_2}= \dfrac{2.25\times 10^{-2}}{2\times 10^{-2}}$

$\Rightarrow n_2 = \dfrac{1.5\times 2\times 10^{-2}}{2\times 10^{-2}}= \dfrac{4}{3}$

A convex mirror of focal length

$10cm$ is shown in figure. A linear object

$AB=5cm$ is placed along the optical axis. Point

$A$ is at distance

$25cm$ from the pole of mirror. The size of image of

$AB$ is:

0%
$2.5cm$
0%
$0.64cm$
0%
$0.36cm$
0%
None of these

When an object is placed at a distance of

$25 \ cm$ from a mirror, the magnification is

$m_1$ . But the magnification becomes

$m_2$ , when the object is moved

$15 \ cm$ farther away with respect to the earlier position. If

$\dfrac{m_1}{m_2}=4$ , then find the focal length of the mirror and what type of mirror it is?

0%
$20 \ cm$ , convex
0%
$20 \ cm$ , concave
0%
$10 \ cm$ , convex
0%
$10 \ cm$ , concave

The second lens in this optical instrument can not :

0%
Cause the light rays to focus closer than they would with the first lens, acting alone
0%
Cause the light rays to focus farther away than they would with the first lens acting alone
0%
Cause the light beam to diverge after refraction from it
0%
Make the beam parallel

An object of size

$7.5\ cm$ is placed in front of a convex mirror of radius of curvature

$25\ cm$ at a distance of

$40\ cm.$ The size of the image should be

0%
$2.3\ cm$
0%
$1.7\ cm$
0%
$1\ cm$
0%
$0.8\ cm$

Explanation

By using

$\dfrac{I}{O}=\dfrac{f}{f-u}$

$\Rightarrow \dfrac{I}{+(7.5)}=\dfrac{(25/2)}{\left(\dfrac{25}{2}\right)-(-40)}\Rightarrow I=1.78\ cm$

CBSE Questions for Class 10 Physics Light Reflection And Refraction Quiz 11 - 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 11 - MCQExams.com

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Practice Class 10 Physics Quiz Questions and Answers

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