Increase the contrast of the image

MCQ Questions for Class 10 Physics Light Reflection And Refraction Quiz 8

An object is placed somewhere in front of concave mirror. The focal length of the mirror is

$10.0 cm$ . The image for this object CANNOT be formed in which of the following locations?

0%
$25.0 cm$ in front of the mirror
0%
$20.0 cm$ in front of the mirror
0%
$15.0 cm$ in front of the mirror
0%
$5.0 cm$ in front of the mirror
0%
$10.0 cm$ behind the mirror

Explanation

Given :

$f = -10$ cm

Let the object is placed at a distance

$x$ from the pole of mirror i.e

$u = -x$ .

Using mirror formula :

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

$\therefore$

$\dfrac{1}{v}+ \dfrac{1}{-x} = \dfrac{1}{-10}$

$\implies v = \dfrac{10x}{10-x}$

For option D :

$v = -5.0$ cm

$\therefore$

$-5.0 = \dfrac{10 x}{10-x}$

$\implies x = -10$ cm (not possible)

Hence option D is correct.

Material

$1$ has an index of refraction of

$1.15$ .
Material

$2$ has an index of refraction of

$2.30$ .
If light passes from air into each of these materials at the same angle of incidence, how will the angle of refraction in material

$1$ compare to the angle of refraction in material

$2$ ?

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The angle of refraction in material $1$ will be half as much as the angle of refraction in material $2$
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The angle of refraction in material $1$ will be greater than the angle of refraction in material $2$
0%
The angle of refraction in material $1$ will be less than the angle of refraction in material $2$ but not necessarily half as much as the angle of refraction in material $2$
0%
The angle of refraction in material $1$ will be less than the angle of refraction in material $2$ but not necessarily twice as much as the angle of refraction in material $2$
0%
The angle of refraction in material $1$ will be one-fourth as much as the angle of refraction in material $2$

Explanation

$n_1$ =

$1.15$

$n_2$ =

$2.30$

Given :

$n_1< n_2$

Refractive index of air

$n_a = 1.00$

Using Snell's law of refraction,

$n_{air} \times sin i = n_{medium} \times sin r$

$\therefore$

$1.00 \times sin i = n_{medium} \times sin r$

$\implies$

$sin r = \dfrac{sin i}{n_{medium}}$

Hence

$r$ is inversely proportional to refractive index

$n$

Thus

$n_1< n_2$

$\implies$

$r_1>r_2$

Above, a light ray is pictured entering a piece of glass from the air.
Which diagram most accurately shows the path the light that passes through the glass will take?

Which of the following are the properties of plane mirror images?

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The image is the same size as the object
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The image is virtual
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The image is laterally inverted
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Image distance is equal to object distance

We want a mirror that will make an object look larger. What combination of image and object distances (from the mirror) will accomplish this?

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Image Distance $3.0 cm$ , Object Distance $3.0 cm$
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Image Distance $2.0 cm$ , Object Distance $3.0 cm$
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Image Distance $3.0 cm$ , Object Distance $5.0 cm$
0%
Image Distance $3.0 cm$ , Object Distance $2.0 cm$
0%
Image Distance $3.0 cm$ , Object Distance $10.0 cm$

Explanation

Magnification by a mirror

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

The image distance

$|u|$ should be less than object distance

$|v|$ so that

$m>1$ i.e. magnified image. Here we don't have to consider the sign of

$u$ and

$v$ as we are not concerned about the nature of the imgae. We are interested only in the size of image.

For option D :

$m = \dfrac{- 3.0}{-2.0} = 1.5$

$\implies m>1$

Thus the data given in option D will make a larger image.

A real object is placed in front of a convex lens of focal length

$f$ at its principal focus. Then the image is formed at

0%
zero
0%
infinity
0%
at a distance $2f$
0%
at a distance $\dfrac f2$

Explanation

The object is placed at focus so

$u=-f$

For lens formula

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

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

$\dfrac 1v = 0$

Hence,

$v=\infty$ , the image is formed at infinity.

In case of concave mirror, the minimum distance between a real object and its real image is :

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$f$
0%
$2f$
0%
$4f$
0%
$zero$

Explanation

Hint :- When object is placed at center of curvature then image is also formed at centre of curvature.

Explanation:-

$\textbf{Step1: Ray diagram}$

$\textbf{Step2: Apply mirror formula}$

Let focal length be f and object is placed at

$u= 2f$

So by mirror formula

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

$v=-2f$ .

The image will form at the object location.

So minimum distance is zero.

$\textbf{Hence option D correct}$

The image distance of an object placed 10 cm in front of a thin lens of focal length +5 cm is at :

0%
6.5 cm
0%
8.0 cm
0%
9.5 cm
0%
10.0 cm

Explanation

Given:

The object distance

$u=-10\ cm$

The focal length of the lens

$f=+5\ cm$

From the lens formula,

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

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

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

$\therefore v=10\ cm$

Thus, the distance of the image from the lens is

$v=10cm$ and it is on the other side of the lens.

A concave mirror gives an image three times as large as the object placed at a distance of 20 cm from it. For the image to be real, the focal length should be.

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10 cm
0%
15 cm
0%
20 cm
0%
30 cm

Explanation

Since the image formed by concave mirror is real ,so it must be an inverted image.

Distance of object from mirror

$u=-20cm$

Since image is inverted then the magnification will be negative ,hence

$m=-3$

Magnification

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

$m=-3=-\dfrac{v}{-20}$

$v=-60cm$

we know for mirror

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

$\dfrac{1}{f}=\dfrac{1}{-60}+\dfrac{1}{-20}$

$\dfrac{1}{f}=\dfrac{-60-20}{60\times20}$ =

$\dfrac{-80}{1200}$ =

$\dfrac{-1}{15}$

$f=-15cm$

An object 1 cm tall is placed 4 cm in front of a mirror. In order to produce an upright image of 3 cm height, one needs a

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convex mirror of radius of curvature 12 cm
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concave mirror of radius of curvature 12 cm
0%
concave mirror of radius of curvature 4 cm
0%
plane mirror of height 12 cm

Explanation

The concave mirror can produce a magnified upright image.

Distance of object from mirror

$u=-4$ cm

Magnification

$m=\dfrac{\text{Image height}}{\text{Object height}}=\dfrac31=-\dfrac{v}{u} = -\dfrac{v}{-4}$

$v=12cm$

Now we know for mirror

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

$\dfrac{1}{f}=\dfrac{1}{12}+\dfrac{1}{-4}$

$f=-6$ cm

Radius of curvature is

$R=2f=12cm$ and mirror is concave because the focal length is negative.

Find the size of the image formed by a spherical mirror from the following data :

$u = -20\ cm$ ,

$f = -15\ cm$ and height of object

$= 1.0\ cm$

0%
$6\ cm$
0%
$5\ cm$
0%
$3\ cm$
0%
$4\ cm$

Explanation

Given:

$u=-20$

$f=-15$

We know

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

$\dfrac{1}{-15}=\dfrac{1}{v}+\dfrac{1}{-20}$

$v=-60cm$

Now the magnification

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

Height of object

$h=1cm$

Height of image

$h_1=|m|\times h=3\times 1=3cm$

A ray of light strikes a glass plate at an angle of

${60}^{o}$ . If the reflected and refracted rays are perpendicular to each other, the refractive index of glass is

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

Explanation

Let the refractive index of the material be

$\mu$ , and the ange of refraction be

$r$ .

Since the reflected ray and refracted ray are perpendicular to each other,

$60^{\circ}+r=90^{\circ}$

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

From snell's law,

$sin i=\mu sinr$

$\implies \mu=\dfrac{sin60^{\circ}}{sin30^{\circ}}$

$=\sqrt{3}$

Focal length of a convex lens is

$20 cm$ and its RI is

$1.5$ . It produces an erect, enlarged image if the distance of the object from the lens is

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$40 cm$
0%
$30 cm$
0%
$15 cm$
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$20 cm$

Explanation

A convex lens forms an erect and enlarged image when the object is placed between focus and the lens i.e. object distance must be less than the focal length of the lens.

Thus object distance must be

$15 cm$ (which is less that

$20 cm$ ).

When rays of light are incident on a glass slab, then the incident ray and emergent ray are _________ to each other.

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Perpendicular
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Parallel
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Opposite
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Concurrent

A glass slab is placed in the path of convergent light. The point of convergence of light :

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

The distance between principal focus and optical center of the lens is ________.

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Diameter
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Focal length
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Principal axis
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Optical center

In the diagram, the correctly marked angles are ______.

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$\angle i$ and $\angle r$
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$\angle i$ and $\angle e$
0%
$\angle r$ and $\angle e$
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$\angle i$ , $\angle r$ and $\angle e$

Explanation

All the angles should be taken from the normal to the ray. So in the figure only

$i$ and

$r$ angles are marked with the normal and hence correct.

Angle

$e$ is marked between the ray and the interface boundary so it is incorrect. Hence option (a) is the answer.

Magnification for mirror

$(m) =$ ______

0%
$\dfrac {v}{u}$
0%
$\dfrac {u}{v}$
0%
$\dfrac {h_{o}}{h_{i}}$
0%
$\dfrac {h_{i}}{h_{o}}$

Explanation

Magnification is ratio of the size of the image

$h_i$ to the size of the object

$h_o$ .

$m=\dfrac{h_i}{h_o} = \dfrac{-v}{u}$ where

$u$ is object distance and

$v$ is image distance.

After refraction of light through a glass slab, incident ray and emergent ray are:

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Perpendicular
0%
Parallel
0%
In a straight line
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None of these

For magnification in spherical mirrors object height is :

0%
Negative.
0%
Positive.
0%
For real images positive.
0%
For virtual images negative.

Which of the following is Snell's law?

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$n_{1}\sin i = \dfrac {\sin r}{n_{2}}$
0%
$\dfrac {n_{1}}{n_{2}} = \dfrac {\sin r}{\sin i}$
0%
$\dfrac {n_{2}}{n_{1}} = \dfrac {\sin r}{\sin i}$
0%
$n_{2}\sin i = constant$

Explanation

Snell's law of refraction states that the product of sine of angle made by the ray in a medium and the refractive index of that medium is constant i.e.

$n_1\times \sin i =$ constant

We get

$n_1 \sin i = n_2 \sin r$

$\implies$

$\dfrac{n_1}{n_2} = \dfrac{\sin r}{\sin i}$

An object is placed 25 cm from a convex lens whose focal length is 10 cm. The image distance is ________ .

0%
50 cm
0%
16.66 cm
0%
6.66 cm
0%
10 cm

Explanation

Given that,

Object distance,

$u=-25cm, Focal\ length, f=10cm$

From lens formula,

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

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

$\implies \dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{25}$

$\implies \dfrac{1}{v}=\dfrac{5-2}{50}=\dfrac{3}{50}$

$\implies v=16.66cm$

Answer-(B)

According to the new cartesian sign convention, the ________ is taken as origin

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Centre of curvature
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Pole of the mirror
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Focus
0%
Any point on the principal axis

Rays of light are entering from glass to glycerine. If refractive indexes of glass and glycerine are respectively 1.5 and 1.47, find the refractive index of glycerine with respect to glass.

0%
0.03
0%
1.02
0%
2.20
0%
0.98

Explanation

Absolute refractive index of glass

$n_1 = 1.5$
Absolute refractive index of glycerine

$n_2 = 1.47$
R.I. of glass relative index

$n_2 = ?$

$n_{21} \dfrac {n_2}{n_1} = \dfrac {1.47}{1.5} = 0.98$
R.I. of glass relative to glycerine is 0.98

Why is a Converging lens so called ?

0%
The rays of light passing through it meet at a point.
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The rays of light passing through it pass without any deviation.
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The rays of light passing through it move away from each other.
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None of these

Which of the following is correct regarding reflection.

0%
Reflection is a phenomenon by which light incident on a surface is sent back into the same medium
0%
Reflection is a phenomenon by which light incident on a surface passes into another medium
0%
Reflection is a phenomenon by which light incident on a surface is scattered
0%
Reflection is a phenomenon by which light incident on a surface is absorbed by the surface

A convex mirror of radius of curvature

$1.6 m$ has an object placed at a distance of

$1m$ from it. The image is formed at a distance of :

0%
$\dfrac{18}{3}m$ behind the mirror
0%
$\dfrac{18}{3}m$ infront of the mirror
0%
$\dfrac{4}{9}m$ behind the mirror
0%
$\dfrac{4}{9}m$ infront of the mirror

Explanation

Given,

$f=\dfrac{1.6}{2}=0.8m$ and

$u=-1m$

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

$\dfrac{1}{v} = \dfrac{1}{0.8} - \dfrac{1}{-1} = \dfrac{10}{8} + 1=\dfrac{9}{4}$

$\Rightarrow v=\dfrac{4}{9}m$

The apparent flattening of the sun at sunset and sunrise is due to

0%
refraction
0%
diffraction
0%
total internal reflection
0%
interference
0%
polarization

If the angle of incidence is i and that of refraction is r. Then the speed of light in the medium to which the light is reflected from air is?

0%
$\displaystyle v=C\frac{\sin i}{\cos r}$
0%
$\displaystyle v=C\frac{\cos r}{\cos i}$
0%
$\displaystyle v=C\frac{\sin r}{\sin i}$
0%
$\displaystyle v=C\frac{\sin i}{\sin r}$

Explanation

Given,

Angle of incidence is

$i$

Angle refraction is

$r$

From Snell's law, refracting index of second medium with respect to first medium

$\mu=\displaystyle\frac{\sin i}{\sin r}$ ......(1)

also

$\mu=\displaystyle\frac{c}{v}$ .........(2)

From above two equations

$\therefore \displaystyle \mu=\frac{c}{v}=\frac{\sin i}{\sin r}$

$\Rightarrow v=c\cdot \left(\displaystyle\frac{\sin r}{\sin i}\right)$ .

Hence option C

If the angle of incidence is twice the angle of refraction in a medium of refractive index

$\mu$ , then the angle of incidence is :

0%
$2\, cos ^{-1}\dfrac{\mu}{2}$
0%
$2\, sin ^{-1}\dfrac{\mu}{2}$
0%
$2\, cos ^{-1} \mu$
0%
$2\, sin ^{-1} \mu$

In optical instruments, the lenses are used to form the image by the phenomenon of ...................................

0%
Reflection
0%
Refraction
0%
Scattering
0%
Diffusion

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

$\dfrac {1}{\sqrt {2}}$ with respect to air. When it is incident on the surface making an angle

$45^{\circ}$ with the surface, which of the following will take place?

0%
Angle of refraction will be $45^{\circ}$
0%
Angle of refraction will be $90^{\circ}$
0%
The ray will be internally reflected
0%
None of the above

Explanation

Given,

$\mu = \dfrac {1}{\sqrt {2}} w.r.t.$ air

$i = 45^{\circ}, r = ?$
Also,

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

$\Rightarrow \dfrac {1}{\sqrt {2}} = \dfrac {\sin 45^{\circ}}{\sin r} \Rightarrow \dfrac {1}{\sqrt {2}} =\dfrac {1/\sqrt {2}}{\sin r}$

$\Rightarrow \sin r = 1$

$\Rightarrow r = 90^{\circ}$
Note that the ray is incident from denser medium and is refracted into a rarer medium air.

A real object is placed at a distance

$f$ from the pole of a convex mirror, in front of the convex minor. If focal length of the mirror is

$f$ , then distance of the image from the pole of the mirror is :

0%
$2\, f$
0%
$\frac{f}{2}$
0%
$4\, f$
0%
$\frac{f}{4}$

Explanation

Here the object distance is

$f$ ,
So

$u=-f$ and

$F=f$

Using the mirror's formula

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

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

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

Convex lens

0%
Shrinks the image
0%
Magnifies the image
0%
Sharpens the image
0%
Increase the contrast of the image

Lens spectacles are made from

0%
Soft Glass
0%
Hard Glass
0%
Pyrex Glass
0%
Flint Glass

You are given two lenses, a converging lens with focal length + 10 cm and a diverging lens with focal length -20 cm. Which of the following would produce a real image that is smaller than the object?

0%
Placing the object 5 cm from the converging lens
0%
Placing the object 15 cm from the converging lens
0%
Placing the object 25 cm from the converging lens
0%
Placing the object 15 cm from the diverging lens

Explanation

When the object is placed beyond the center of curvature of a converging lens, a real and diminished image is formed between the focus and the center of curvature.

The focal length of the converging lens is

$+10\ cm$

The radius of curvature of the converging lens is

$2f=+20\ cm$

Thus, the object must be placed beyond

$20\ cm$ or at

$25\ cm$ .

1443848_769389_ans_f6785d24f5664de4960fd14aaaa540a7.png

How will you design a shaving mirror assuming that a person keeps it

$10cm$ from his face and view the magnified image of the face at the closest comfortable distance of

$25cm$ . The radius of curvature of the mirror would then be:

0%
$24cm$
0%
$-33.3cm$
0%
$-24cm$
0%
$30cm$

Explanation

Given:

A person keeps a mirror at a distance of

$10cm$ from his face and view the magnified image of the face at the closest comfortable distance of

$25cm.$

Since magnified image on other side of mirror is produced it is clearly a concave mirror with case of object between POLE and FOCUS of mirror.

To find the radius of curvature of the mirror

Solution:

From given criteria,

Object distance,

$u=-10cm$

Image distance,

$v=25cm$ refer above fig.

Applying the mirror formula, we get

$\dfrac 1f=\dfrac 1v+\dfrac 1u\\\implies \dfrac 1f=\dfrac 1{25}+\dfrac 1{(-10)}\\\implies \dfrac 1f=\dfrac {-3}{50}\\\implies f=\frac{-50}{3}cm$

And Radius of curvature,

$R=2f=2\times \frac{-50}{3}=-33.3cm$

Hence the radius of curvature of the mirror will be

$-33.3 cm.$

The magnification of the image when an object is placed at a distance x from the principle focus of a mirror of focal length f is?

0%
$\displaystyle\frac{x}{f}$
0%
$1+f/x$
0%
$\displaystyle\frac{f}{x}$
0%
$1-\displaystyle\frac{f}{x}$

Explanation

$\textbf{Step 1: Sign Convention and position of Object [Refer fig.]}$

Pole of the mirror is taken as origin, and all distances are taken from pole.

Sign convention

$\rightarrow$ Direction of incident ray is +ve, i.e. rays coming from object to mirror, rightwards here.

As Object

$(O)$ is leftwards, So object distance

$u= -(x + f)$

As focus is also to the left of the pole, So focal length

$=-f$

$\textbf{Step 2: From mirror formula}$

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

$\dfrac{1}{-f} = \dfrac{1}{v} + \dfrac{1}{-(x + f)}$

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

$\dfrac{1}{v} = \dfrac{f - x - f}{f (x + f)}$

$v = - \dfrac{f(x + f)}{x}$

$\textbf{Step 2: Magnification of mirror}$

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

$m = \dfrac{\dfrac{-f (x + f)}{x}}{- (x + f)}$

$=\ \ -\dfrac{f}{x}$

$\therefore\ \ \ |m| = \dfrac{f}{x}$

Hence, Option

$C$ is correct

2110178_692897_ans_bedc9d0e8aa143e39f852442ff69bcb6.png

The refractive index of glass with respect to air is

$\dfrac{3}{2}$ and that of water with respect to air is

$\dfrac{3}{2}$ . What is the refractive index of glass with respect to water?

0%
$9 : 8$
0%
$1 : 2$
0%
$2 : 1$
0%
$8 : 9$

Explanation

Given,

Refractive index of glass=

$\dfrac{3}{2}$

Refractive index of water=

$\dfrac{4}{3}$

Refractive Index of glass with respect to water=

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

$\dfrac{9}{8}$

The phenomenon of bending of light at the surface of separation of two media is called :

0%
Refraction of light
0%
Reflection of light
0%
Deflection of light
0%
Absorption of light

Which among these is the angle of incidence?

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

Magnification produced by a convex mirror is

$\dfrac{1}{3}$ , then distance of the object from mirror is

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

Explanation

$\dfrac{1}{f} = \dfrac{1}{v}-\dfrac{1}{u}$ , multiply using by u

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

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

$\Rightarrow \dfrac{v}{u} = \dfrac{f}{f+u}\Rightarrow m|\dfrac{-v}{u}| = \dfrac{f}{f+u}$

given that

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

$\dfrac{1}{3} = \dfrac{f}{f+u}\Rightarrow f+u = 3f$

$\Rightarrow \boxed{u = 2f}$

1181988_872952_ans_6f53e156a2cc4a3d8dff2f0ed1976e97.jpg

A light ray passes through a glass slab. Which of the following diagrams shows the correct path of the ray?

A plane mirror is placed along the x-axis facing negative y-axis. The mirror is fixed. A point object is moving with

$3\hat{i} \, + \, 4\hat{j}$ in front of the plane mirror. The relative velocity of image with respect to its object is

0%
$-8\hat{j}$
0%
$8\hat{j}$
0%
$3\hat{i} \, - \, 4\hat{j}$
0%
$-6\hat{i}$

Explanation

Velocity of object,

$\bar{v}_{ob} = 3 \hat{i} + 4 \hat{j}$
Velocity of image

$\bar{v}_{image} = 3 \hat{i} - 4 \hat{j}$
Relative velocity of image with respect to its object

$\bar{v}_{rel} \, = \, \bar{v}_{image} \, - \, \bar{v}_{ob} \, = \, (3 \hat{i} \, - \, 4 \hat{j}) \, - \, (3 \hat{i} \, + \, 4 \hat{j}) = -8 \hat{j}$

An object 2 cm high is placed at a distance of 16 cm from a concave mirror, which produces a real image 3 cm high. What is the focal length of the mirror ?

0%
-9.6 cm
0%
-3.6 cm
0%
-6.3 cm
0%
-8.3 cm

Explanation

Here

$h_1$ = 2 cm, u = -16 cm,

$h_2$ = -3 cm (Since image is real and inverted.)

$\because \, m \, = \, \dfrac{h_2}{h_1} \, = \, - \, \dfrac{v}{u} \, \therefore \, v \, = \, \dfrac{-h_2}{h_1} u \, = \,\dfrac{3}{2} \, \times \, (-16) \, = \, -24 \, cm$

$\dfrac{1}{f} \, = \, \dfrac{1}{v} \, + \, \dfrac{1}{u} \, =\, -\dfrac{1}{24} \, - \, \dfrac{1}{16} \, = \, \dfrac{-2 \, -3}{48} \, = \, \dfrac{-5}{48};f \, = \, \dfrac{-48}{5} \, = \, -9.6 \, cm$

A rod of length

$10 \ cm$ lies along the principal axis of a concave mirror of focal length

$10 \ cm$ in such a way that its end closer to the pole is

$20 \ cm$ away from the mirror. The length of the image is

0%
$10 \ cm$
0%
$15 \ cm$
0%
$2.5 \ cm$
0%
$5 \ cm$

Explanation

The position of image of nearer end to mirror can be found out using mirror equation,

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

$\implies \dfrac{1}{v_1}=\dfrac{1}{-20}-\dfrac{1}{-10}$

$\implies v_2=-20$

Similarly position of image of farther end to mirror can be found out as

$=\dfrac{1}{v_2}=\dfrac{1}{-(10+20)}-\dfrac{1}{-10}$

$\implies v_2=-15$

Hence the length of the image

$=20cm-15cm=5cm$

1490859_952199_ans_5128ac49dfa84613b47c5c43700b3407.png

A convergent beam of light passes through a diverging lens of focal length

$0\cdot2\ m$ and comes to focus

$0\cdot3\ m$ behind the lens. The position of the point at which the beam would converge in the absence of the lens is

0%
$0.12\ m$
0%
$0.6\ m$
0%
$0.3\ m$
0%
$0.15\ m$

Explanation

Here, f = -0.2 m, v = +0.3 m
The lens formula

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

$\Rightarrow \, \dfrac{1}{u} \, = \, \dfrac{1}{v} \, - \, \dfrac{1}{f} \, = \, \dfrac{1}{0.3} \, + \, \dfrac{1}{0.2} \, = \, \dfrac{0.5}{0.006}$

$u \, = \, \dfrac{0.006}{0.5} \, = \, 0.12 \, m$

For which of the pairs of

$u$ and

$f$ for a mirror image is smaller in size

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$u=-10\ cm$ , $f=20\ cm$
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$u=-20\ cm$ , $f=-30\ cm$
0%
$u=-45\ cm$ , $f=-10\ cm$
0%
$u=-60\ cm$ , $f=30\ cm$

An object is placed at $20\ cm$ form a convex mirror of focal length $10\ cm.$ The image formed by a mirror is

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Real and at $20cm$ from the mirror
0%
Virtual and at $20cm$ from the mirror
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Virtual and at $20/3\ cm$ from the mirror
0%
Real and at $20/3\ cm$ from the mirror

Explanation

$u=-20\ cm, f=+10\ cm$ a\so

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

$\Rightarrow \dfrac{1}{+10}=\dfrac{1}{v}+\dfrac{1}{(-20)}\Rightarrow v=\dfrac{20}{3}\ cm;$ virtual image.

A square wire of side 1 cm is placed perpendicular to the principle axis of a concave mirror of focal length 15 cm at a distance of 20 cm. The area enclosed by the image of the wire is:

0%
4 $cm^2$
0%
6 $cm^2$
0%
2 $cm^2$
0%
9 $cm^2$

CBSE Questions for Class 10 Physics Light Reflection And Refraction Quiz 8 - 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 8 - MCQExams.com

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Practice Class 10 Physics Quiz Questions and Answers

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