MCQ Questions for Class 10 Physics Light Reflection And Refraction Quiz 3

What is the distance

$PA$ called?

0%
Equivalent distance
0%
Equivalent radius
0%
Radius of curvature
0%
Center of curvature

Explanation

When the image forms on the object in a concave mirror, then the object as well as the image is situated at the center of the curvature of the concave mirror. Hence, the distance

$PA$ is called as Radius of Curvature of the concave mirror.

Relation between radius of curveture(R) and focal length(f)-

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

$PB = \dfrac{PA}{2}$

A ray of light travelling in air is incident on the plane of a transparent medium. The angle of incident is

$45^0$ and that of refraction is

$30^0$ . The refractive index of the medium with respect to air is:

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

Explanation

Given:

Angle of incidence

$\angle i=45^0$
Angle of refraction

$\angle r=30^0$

Refractive index of transparent medium w.r.t. air is

$\mu_{ma}=\dfrac{\mu_m}{\mu_a}=\dfrac{\mu_m}{1}=\mu_m$

By Snell's Law,

$\displaystyle \mu_m=\dfrac {sin \ i}{sin \ r}=\dfrac {sin \ 45^0}{sin \ 30^0}=\cfrac {\dfrac {1}{\sqrt 2}}{\dfrac {1}{2}}=\sqrt 2$

Find the refractive index of glass with respect to water. The refractive indices of glass and air with respect to air are

$3/2$ and

$4/3$ respectively.

0%
$2/1$
0%
$1/2$
0%
$9/8$
0%
$8/9$

Explanation

Refractive index of medium

$1$ with respect to medium

$2$ ,

$^1\mu_2=\dfrac {\mu_2}{\mu_1}$

Refractive index of glass with respect to air,

$^g\mu_a=\dfrac {3}{2}$

Refractive index of water with respect to air,

$^w\mu_a=\dfrac {4}{3}$

$^g\mu_a=\dfrac {\mu^a}{\mu^g}$

$^w\mu_a=\dfrac {\mu^a}{\mu^g}$

$\therefore \text{refractive index of glass with respect to water,}\ \displaystyle ^g\mu_w=\dfrac {\mu_w}{\mu_a}\times \dfrac {\mu_a}{\mu_g}=\dfrac {^g\mu_a}{^w\mu_a}=\dfrac {3}{2}\times \dfrac {3}{4}=\dfrac {9}{8}$

In case of a virtual and erect image, the magnification of a mirror is :

0%
Positive
0%
Negative
0%
Unity
0%
Infinity

Explanation

For any erect image, the magnification of a mirror is positive.
We know,
Magnification (M)

$= \dfrac{\text{height of image}(h_i)}{\text{height of object}(h_o)}$
The erect image formed same side of object to the principle axis.

So, In Sign convention, Sign of object height and image height will be same.

Hence Magnification will be positive.

If the magnification of a body of size 1 m is 2, what is the size of the image?

0%
1m
0%
2m
0%
3m
0%
4m

Explanation

$m=2=\dfrac {h_i}{h_o}=\dfrac {h_i}{1}\Rightarrow h_1=2\times 1\ m$
So,

$h_1=2\ m$

Formula of focal length in convex lens is

0%
$\displaystyle f = \frac{u+v}{u-v}$
0%
$\displaystyle f = \frac{u\times v}{u-v}$
0%
$\displaystyle f = \frac{u-v}{u+v}$
0%
$\displaystyle f = \frac{u+v}{u+v}$

Explanation

$\begin{array}{l}\text { Focal length of lens, from lens formula : } \\\qquad \begin{array}{l}\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \\ \dfrac{1}{f}=\dfrac{u-v}{v \times u} \\ f=\dfrac{v \times u}{u-v}\end{array}\end{array}$

A light ray passing from crown glass to flint glass, then the speed of light is:

0%
smaller in crown glass than in flint glass
0%
larger in crown glass than in flint glass
0%
same in crown glass and in flint glass
0%
zero in crown glass and flint glass

The rays parallel and close to the principal axis are called:

0%
Converging rays
0%
Diverging rays
0%
Coherent rays
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Paraxial rays

Ratio of the size of the image to the size of the object is known as:

0%
Focal plane
0%
Transformation ratio
0%
Efficiency
0%
Magnification

Explanation

Ratio of the size of the image to the size of the object is known as magnification. It is given by

$m = \dfrac{h_{i}}{h_{o}}$

where

$m \rightarrow$ magnification

$h_{i} \rightarrow$ height of image

$h_{O} \rightarrow$ height of object

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

0%
focal length
0%
radius of curvature
0%
principal section
0%
none of these

The centre of curvature of a ______ mirror lies behind the mirror.

0%
Convex
0%
Concave
0%
Both $A$ and $B$
0%
None of these

The central point of a spherical mirror is called:

0%
Pole
0%
Centre of sphere
0%
Centre of curvature
0%
None of these

The unit of magnification is:

0%
m
0%
$m^2$
0%
$m^{-1}$
0%
None of these

Explanation

Magnification is define as ratio of image height to object height.

$M=\dfrac{h_i}{h_o}$

Units, Unit of height is 'meter (m)'

$[M]=\dfrac{[h_i]}{[h_o]}=\dfrac mm$

Hence dimensionless

Option B

The bending of light as it passes from one medium into another is commonly known as

0%
Reflection
0%
Refraction
0%
Scattering
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Dispersion

Two points such that each focus for the rays proceedings from the other, are called :

0%
Centres of curvature
0%
Conjugate foci
0%
Converging foci
0%
None of these

The angle which subtends the periphery of the spherical mirror at the centre of curvature is called:

0%
angular aperture
0%
glancing angle
0%
critical angle
0%
none of these

The centre of curvature of a ___________ mirror is in front of it.

0%
convex
0%
concave
0%
convex or concave
0%
none of these

Which of the following quantity does not have any unit?

0%
Velocity of light
0%
Light year
0%
Magnification
0%
Power of a lens

In a concave mirror, an object is placed a distance

$d_1$ from the focus and the real image is formed at a distance

$d_2$ from the focus. Then the focal length of the mirror is

0%
$\sqrt {d_1d_2}$
0%
$d_1d_2$
0%
$\frac {d_1d_2}{2}$
0%
$\sqrt {d_1/d_2}$

Explanation

Given: In a concave mirror, an object is placed a distance

$d_1$ from the focus and the real image is formed at a distance

$d_2$ from the focus.

To find the focal length of the mirror

Solution:

Let the focal length of the concave mirror be

$f$

And as the image formed is real, the object distance and image distance have same sign convention.

According to the given criteria,

Object distance,

$u=f+d_1$

Image distance,

$v=f+d_2$

Now by applying mirror formula, we get

$\dfrac 1f=\dfrac 1v+\dfrac 1u\\\implies \dfrac 1f=\dfrac {v+u}{vu}\\\implies vu=f(v+u)$

Substituting the values of

$v$ and

$u$ , we get

$(f+d_2)(f+d_1)=f(f+d_2+f+d_1)\\\implies f^2+fd_1+fd_2+d_1d_2=2f^2+fd_1+fd_2\\\implies 2f^2-f^2=d_1d_2\\\implies f^2=d_1d_2\\\implies f=\sqrt{d_1d_2}$

is the required focal length of the concave mirror.

A glass slab is placed in the path of a beam of convergent light; the point of convergence of light

0%
Moves towards the glass slab
0%
Moves away from the glass slab
0%
Remains at the same point
0%
Undergoes a lateral shift

A lens which is thinner at the middle and thicker at the edges is called a :

0%
Convex lens
0%
Concave lens
0%
Cylindrical lens
0%
None of these

The focal length of a lens is 0.1 m. Then the lens must be :

0%
Convex
0%
Concave
0%
Cylindrical
0%
None of these

An object is placed on the principal axis of a concave mirror at a distance of

$60\ cm$ . If the focal length of the concave mirror is

$40\ cm$ then the magnification obtained is equal to:

0%
$-2$
0%
$-1.5$
0%
$-3$
0%
$-1$

Explanation

Given, Concave mirror

object distance,

$u = -60\ cm$

focal length,

$f = -40\ cm$

image distance,

$v$

Using mirror formula,

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

$\dfrac{1}{-40} = \dfrac{1}{-60} + \dfrac{1}{v}$

$\dfrac{1}{v} = \dfrac{1}{-40} + \dfrac{1}{60} = \dfrac{-3 + 2}{120}$

$v = -120\ cm$

Magnification,

$m$

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

$m= \dfrac{-\left(-120\right)}{-60}$

$m = -2$

A lens which is thicker in the middle and thinner at the edges is called a :

0%
Convex lens
0%
Concave lens
0%
Cylindrical lens
0%
None of these

The expression for the magnification of a spherical mirror in the terms of focal length (f) and the distance of the object from mirror (u) is

0%
$\frac{-f}{u-f}$
0%
$\frac{f}{u+f}$
0%
$\frac{-f}{u+f}$
0%
$\frac{f}{u-f}$

Explanation

Equation of spherical mirror is

$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$ , where v is the image distance.

solving,

Replacing

$v$ with

$v=-mu$ ,

$\dfrac{1}{f} = \dfrac{1}{u} (1 - \dfrac{1}{m} )$

$u = f (\dfrac{1}{m} +1)$ where m = magnification

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

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

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

The constant ratio of the sine of angle of incidence to the sine of angle of refraction is also known as:

0%
Snell's law
0%
Optical density
0%
Relative density
0%
None of these

When light travels from one medium into another it suffers

0%
Reflection
0%
Refraction
0%
Dispersion
0%
None of these

A ray of light is incident on a medium with angle of incidence i and refracted into a second medium with angle of refraction

$r$ . The graph of

$\sin(i)$ Vs

$\sin(r)$ is as shown in the figure. Then the velocity of light in the first medium is ................. times the velocity of light in the second medium.

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

Explanation

Given

Ray of light is refracted when it is passed from one medium to another

Solution

Let refractive index of medium 1 is

$\mu_{1}$

similarly refractive index of medium 2 is

$\mu_{2}$

Using Snell law

$\mu_{1}\sin i =\mu_{2}\sin r$

$\dfrac{\mu_{1}}{\mu_{2}}=\dfrac{\sin r}{\sin i}$

Slope of graph

$=\tan 30 =\dfrac{\sin r}{\sin i}$

$\dfrac{\mu_{1}}{\mu_{2}}=3^{-1/2}$

We know that velocity of light in a medium is inversely proportional to refractive index, ie

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

$\dfrac{\mu_{1}}{\mu_{2}}=\dfrac{v_{2}}{v_{1}}=3^{-1/2}$

$v_{1}=3^{1/2} \times v_{2}$

The correct option is A

Ram and Shyam observe a virtual image formed by a real object placed a distance

$\dfrac{f}{2}$ in front of a convex lens of focal length

$f$ . Now they want to use a concave lens having a focal length f so that a virtual image is formed at the same distance from the lens as before. Where should the object be kept relative to the concave lens

0%
The object should be at innity
0%
Object should be at a distance of $f$
0%
Object should be at $f/2$
0%
Object should be at $2f$

Explanation

$v= -\dfrac{2}{f}$ ;

$f= -f$

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

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

$u= -f$

It should be placed at a distance

$f$ from the lens
option

$B$ is correct

Fig shows an object and its image formed by a thin lens. Then the nature and focal length of lens is

0%
$f=4.8$ cm converging lens
0%
$f=-4.8$ cm diverging lens
0%
$f=2.18$ cm converging lens
0%
$f=-2.18$ cm diverging lens

Explanation

From figure,

$u = -(3+5) = -8$ cm,

$v = -3$ cm

Using lens formula,

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

$\therefore$

$\dfrac{1}{-3} - \dfrac{1}{-8} = \dfrac{1}{f}$

$\implies f = -4.8$ cm

As the focal length of the lens comes out to be negative, thus its a diverging lens (concave lens).

If an object is placed at a distance of 20cm from the pole of a concave mirror, the magnification of its real image isIf the object is moved away from the mirror by 10cm, then the magnification is -1.

0%
True
0%
False

Explanation

Now magnification is given by:

$M=\frac{-v}{u}$

And mirror formula

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

Multiply by $u$

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

$\frac{u}{v}$ + $1$ = $\frac{u}{f}$ ------------ $(1)$

We know $M=\frac{-v}{u}$

$\frac{1}{M}=\frac{-u}{v}$

$\frac{u}{v}=\frac{-1}{M}$

Putting in $(1)$

$\frac{-1}{M}$ + $1$ = $\frac{u}{f}$

$\frac{-1}{M}$ = $\frac{u-f}{f}$

$M=\frac{f}{f-u}$

Now $u=-20$

$M= \frac{f}{f-(-20)}$ and real image means M is negative

$-3= \frac{f}{f+20}$

$-3f-60=f$

$f=-15 cm$

$M= \frac{-15}{-15+30}$

$M= -1$

The image formed by a plane mirror may be -

0%
Real
0%
Virtual
0%
Larger
0%
None of above

A convex lens forms an image of an object placed

$20 cm$ away from it at a distance of

$20 cm$ on the other side of the lens. If the object moves

$5 cm$ towards the lens, then the image will be :

0%
$5 cm$ toward the lens
0%
$5 cm$ away from the lens
0%
$10 cm$ toward the lens
0%
$10 cm$ away from the lens

Explanation

Given,

$u= -20 \space cm$ ;

$v= +20 \space cm$

From lens formula,

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

$\dfrac{1}{20}-\dfrac{-1}{20}=\dfrac{1}{f}$

$f= +10 \space cm$

Now object is moved by

$5 \ cm$ towards the lens
So now,

$u= -15 \space cm$ ;

$f= +10 \space cm$

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

$\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{10}$

$v= +30 \space cm$

So the image has moved

$(30-20)=10 \space cm$

Image is moved

$10 \ cm$ away from the lens.
option

$D$ is correct

A converging beam of rays is incident on a diverging thin lens. Having passed through the lens the ray intersect at a point 15 cm from the lens. If lens is removed the point where the rays meet will move 5 cm closer towards the mounting that holds the lens. The focal length of lens is

0%
$f=10 cm$
0%
$f=15 cm$
0%
$f=30 cm$
0%
$f=40 cm$

Explanation

$v= +15$ ;

$u= +10$

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

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

$f= -30$
option

$C$ is correct

The cause of mirage formation in desert areas is

0%
The refractive index of atmosphere decreases with decrease in height
0%
The refractive index of atmosphere increases with decrease in height
0%
The refractive index of atmosphere does not change with height
0%
Scattering

Explanation

The reason for mirage formation is "The refractive index of atmosphere increases with decrease in height" and hence the rays get totally refracted and forms a mirage.

As we go down in the atmosphere, the air gets denser and hence the velocity in that medium decreases. Since refractive index

$\mu = \dfrac{c}{v}$ . Hence, the refractive index increases as we decrease height.

Which of the following (referred to a spherical mirror) do (does) not depend on whether the rays are paraxial or not?

0%
Pole
0%
Focus
0%
Radius of curvature
0%
Principal axis

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

0%
reflection of light
0%
refraction of light
0%
scattering of light
0%
dispersion of light

Light is incident on a glass block as shown in Fig. If

$\theta_{1}$ is increased slightly, what happens to

$\theta_{2}$ ?

0%
$\theta_{2}$ also increases slightly
0%
$\theta_{2}$ is unchanged
0%
$\theta_{2}$ decreases slightly
0%
$\theta_{2}$ changes abruptly, since the ray experience total internal reflection

An object kept on the principal axis and infront of a spherical mirror, is moved along the axis itself. Its lateral magnification m is measured, and plotted versus object distance |u| for a range of u, as shown in fig. The magnification of the object when it is placed at a distance 20 cm in front of the mirror is:

0%
$-1$
0%
$1$
0%
$8$
0%
$20$

Explanation

We need to find the focal length

$f$ of the mirror. From graph we can notice that when object distance u is -5 cm then the lateral magnification is 2.
As magnification =

$-\dfrac{\text{Image distance}}{\text{Object distance}}$ .
So the image distance:

$v =-2\times 5=-10\ cm$ .
From the mirror equation,

$f=-10\ cm$ .
Now when

$u$ is

$-20\ cm$ , using the mirror equation,

$v = -20\ cm$ , or the image is at the center of curvature of the spherical mirror.
So the magnification here is

$=-\frac{(-20)}{(-20)}=-1$

An object is kept at

$15\ cm$ from a convex mirror of focal length

$25\ cm$ . What is the magnification?

0%
$4/9$
0%
$5/8$
0%
$9/4$
0%
$8/5$

Explanation

Given, (Convex mirror)

Object distance,

$u=-15\ cm$

Focal length,

$f=25\ cm$

Image distance,

$v$

Using mirror formula,

$\dfrac1f=\dfrac1v+\dfrac1u$

$\dfrac1v=\dfrac1f-\dfrac1u$

$\dfrac{1}{v}=\dfrac{1}{25}-\dfrac{1}{-15}$

$\dfrac{1}{v}=\dfrac{3+5}{75}$

$v=\dfrac{75}{8}\ cm$

Magnification,

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

$m=\dfrac{-75}{8\times (-15)}$

$m=\dfrac58$

A ray of light is incident on a medium with angle of incidence

$i$ and refracted into a second medium with angle of refraction

$r$ . The graph of

$sin(i)\ vs\ sin(r)$ is as shown in fig. Then, the velocity of light in the first medium in

$n$ times the velocity of light in the second medium. What should be the value of

$n$ ?

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

Explanation

Using Snell's law,

$\dfrac{\mu_{2}}{\mu_{1}}=\dfrac{sini}{sinr}=tan60^o=\sqrt{3}$

We know,

$\dfrac{v_{1}}{v_{2}}=\dfrac{\mu_{2}}{\mu_{1}}=n=\sqrt{3}$

option

$A$ is correct

The diagram alongside shows the refraction of a ray of light from air to a liquid. The refractive index of liquid with respect to air is :

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

Explanation

According to Snell's law when light goes from rarer to denser medium ,

$n_2 \sin r = n_1 \sin i$ ...............

$(1)$

where,

$n_2$ is refractive index of denser medium(liquid),

$n_1$ is refractive index of rarer medium(air).

Angle of incidence with the normal(dashed line in the figure ) =

$i$

Angle of refraction with the normal =

$r$

According to the given figure,

$i = 60^o$ and

$r = 45^o$

Define

$n = \dfrac {n_2}{n_1}$ , which is refractive index of liquid with respect to air.

Using equation

$(1)$ we get,

$n= \dfrac{\sin 60^o}{\sin 45^o} =\dfrac{\dfrac{\sqrt3}{2}}{\dfrac{1}{\sqrt{2}}}$

$n = \sqrt{\dfrac{3}{2}}$

Option C is correct.

Find the magnification and nature of the image when point

$O$ is

$30\ cm$ behind the mirror.

0%
$2$ (Virtual, Inverted)
0%
$3$ (Real, Inverted)
0%
$5$ (Real, Erect)
0%
$2$ (Virtual, Erect)

Explanation

For this situation, object will be virtual as shown in the figure.

Here,

$u = +10\ cm$ and

$f = +20\ cm$ .

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

So,

$\dfrac {1}{v} + \dfrac {1}{+10} = \dfrac {1}{+20} \Rightarrow i.e., v = -20\ cm$

i.e., the image will be at a distance of

$20\ cm$ in front of the mirror and will be real, erect and enlarged with

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

$m = \frac{-(-20)}{10} = +2$ .

(b.) For this situation also, object will be virtual as shown in the figure.

Here,

$u = +30\ cm$

And

$f = +20\ cm$

$\dfrac {1}{v} + \dfrac {1}{+30} = \dfrac {1}{+20} i.e., v = +60\ cm$

i.e., the image will be at a distance of

$60\ cm$ behind the mirror and will be virtual, inverted, and enlarged with

$m = (-60/30) = -2$ .

1818576_160961_ans_bc906e17ca924aaaa9e328b897b24e5e.jpg

A beam of light converges towards a point O, behind a convex mirror of focal length 20 cm. Find the magnification and nature of the image when point O is 30 cm behind the mirror.

0%
2 (virtual, inverted)
0%
3 (real, inverted)
0%
3, (virtual, enlarged)
0%
+1 (real, enlarged)

Explanation

$u=30$ ;

$f=20$

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

$\dfrac{1}{v}+\dfrac{1}{30}=\dfrac{1}{20}$

$v= 60$

Image is virtual (v>0)

Magnification is

$-\dfrac{v}{u}= -\dfrac{60}{30}= -2$ (<0) hence it is inverted.

The diagram alongside shows the refraction of a ray of light from air to a liquid. Angle of incidence is :

0%
$30^o$
0%
$45^o$
0%
$90^o$
0%
$60^o$

Explanation

Angle of incidence is the angle made between the incident ray and the normal. The angle between the incident ray and surface is

$30^o$ . Hence, angle between normal and incident ray will be 90-30=

$60^o$

The diagram alongside shows the refraction of a ray of light from air to a liquid. Angle of refraction is:

0%
$60^o$
0%
$15^o$
0%
$30^o$
0%
$45^o$

Explanation

Angle of refraction is the angle made between the refracted ray and the normal. The angle between the refracted ray and surface is

$45^o$ . Hence, angle between normal and refracted ray will be 90-45=

$45^o$

The incident ray, the _______ ray and the normal lie in the same plane during refraction.

0%
reflected
0%
refracted
0%
diffused
0%
blocked

In the given figure, name the ray which represents the correct path of light while emerging out through a glass blocks :

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

A ray of light from air suffers partial reflection and refraction at the boundary of water as shown in the figure. In Fig, which of the ray is the correct refracted ray?

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

Explanation

When light travels from a rarer to denser medium, the refracted ray bends towards the normal.
According to Snell's Law:

$\dfrac{n_2}{n_1}=\dfrac{sin\theta_1}{sin\theta_2}$
Refractive index is higher in a denser medium hence angle of refraction in denser medium should be lesser than angle of incidence. Thus, option B.

A ray of light is incident at the glass-water interface at an angle i. It merges finally parallel to the surface of water. Then, the value of

$\mu_g$ would be :

0%
$(4/3) sin i$
0%
$1/ sin i$
0%
$4/3$
0%
$1$

Explanation

Applying Snell's law for glass and water,

$\mu_{g}sin\ i=\mu_{w}sin\ r . . . . . . . . . (i)$

Applying Snell's law for water and air,

$\mu_{w}sin\ r=\mu_{a}sin\ 90^0 . . . . . . (ii)$

Using (i) and (ii), we can apply Snell's Law for glass and air

$\mu_{g}sin\ i=\mu_{a}sin\ 90^0$

$\implies \mu_{g}sin\ i= 1 \times 1$

$\implies \mu_{g}=\dfrac{1}{sin \ i}$

Option

$B$ is correct.

CBSE Questions for Class 10 Physics Light Reflection And Refraction Quiz 3 - 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 3 - MCQExams.com

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Practice Class 10 Physics Quiz Questions and Answers

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