Ray Optics and Optical Test

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Ray Optics and Optical Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

The relation among u, v and f for a mirror is

A
$f = uv/(u+v)$

B
$v = fu/(u+f)$

C
$u = fv/(f+v)$

D
All of these

Solution

According to mirror equation.
$\cfrac{1}{f} = \cfrac{1}{u}+ \cfrac{1}{v}$
$\cfrac{1}{f} = \cfrac{u+v}{uv}$
$f = \cfrac{uv}{u+v}$

Similarly,
$v = \cfrac{fu}{u-f}$
$u = \cfrac{fv}{v-f}$
Question 2
1 / -0

While a moving picture is being screened, a boy introduced a glass slab thickness $3\ cm$ and refractive index $1.5$ between the projector and the screen. In order to have a clear picture on the screen, the screen show to be moved through a distance of :

A
$1\ cm$ away

B
$1\ cm$ nearer

C
$2\ cm$ away

D
$3\ cm$ away

Solution

Refractive index of slab, $n_{slab} = 1.5$

Refractive index of air , $n_{air} = 1$

Thickness of slab, $t = 3 \ cm$

Shift due to slab $= t(1-\dfrac{n_{air}}{n_{slab}}) = 3(1-\dfrac{1}{1.5}) = 1\ cm$

So, the image produced by the projector gets shifted by $1 \ cm$ away from the slab due to refraction. Thus the screen is now has to be shifted by $1 \ cm$ away.
Question 3
1 / -0

In the given figure a Plane concave lens is placed on paper on which a flower is drawn. How far above its actual position does the flower appear to be?

A
$10cm$

B
$15cm$

C
$50cm$

D
none of these

Solution

Since apparent shift = $t\left( \mu -1 \right)$
here t is thickness of glass slab $\mu$ is Refractive index of glass
hence $\delta =20\left( 3/2-1 \right)$
$\delta =10cm$
Question 4
1 / -0

A travelling microscope is focussed on an ink dot. When a glass slab $(n = 1.5)$ of thickness $9\ cm$ is introduced on the dot, the travelling microscope has to be moved by

A
$3\ cm$ upwards

B
$5\ cm$ upwards

C
$3\ cm$ downwards

D
$5\ cm$ downwards

Solution

Refractive index of the glass slab $n = 1.5$
Thickness of slab $t =9 \ cm$
Shift in the position of microscope $x = t\bigg(1-\dfrac{1}{n}\bigg)$
$\therefore$ $x = 9\bigg(1-\dfrac{1}{1.5}\bigg) = 3 \ cm$ upwards
Question 5
1 / -0

The construction of Worlds largest telescope is started in which country?

A
America

B
Denmark

C
Japan

D
Chile

Solution
Question 6
1 / -0

The ray diagram could be correct

A
If ${ n }_{ 1 }={ n }_{ 2 }={ n }_{ g }$

B
If ${ n }_{ 1 }={ n }_{ 2 }$ and ${ n }_{ 1 }<{ n }_{ g }$

C
If ${ n }_{ 1 }={ n }_{ 2 }$ and ${ n }_{ 1 }> { n }_{ g }$

D
$under\ no\ circumstances$

Solution

It is to be remembered that to make a lens nature opposite i.e. from convergent to divergent we should start using it in denser medium or if it is in dense medium then use it in rarer medium.

Hence, ${ n }_{ 1 }={ n }_{ 2 }\quad \& \quad { n }_{ 1 }>{ n }_{ g }$
due to surrounding medium.
Question 7
1 / -0

The relation among $u,v$ and $f$ for a mirror is:

A
$f=uv(u+v)$

B
$v=fu(u+f)$

C
$u=fv(f+v)$

D
None of these

Solution

from mirror formula
$\dfrac { 1 }{ v } +\dfrac { 1 }{ u } =\dfrac { 1 }{ f }$
$\Rightarrow \quad \boxed { f=\dfrac { uv }{ u+v } }$
and
$1+\dfrac { v }{ u } =\dfrac { v }{ f } \Rightarrow \boxed { u=\dfrac { fv }{ f-v } }$
Question 8
1 / -0

A monochromatic ray of light is incident on the surface of a glass slab of thickness $8\ cm$ , making and angle $30^{\circ}$ with the surface. The lateral displacement of the ray of light will be $(\mu_{g} = \sqrt {3})$ .

A
$4.62\ cm$

B
$5.62\ cm$

C
$3.62\ cm$

D
$8\ cm$

Solution

$\begin{array}{l}\mu_{1} \sin i=\mu_{2}\sin\gamma\\\text { (1) }\sin 60=\sqrt{3}\sin\gamma\\\qquad\begin{aligned}\sin\gamma=\dfrac{1}{2} & \text { so consider }\triangle O H A\\\gamma=30^{\circ} &\end{aligned}\end{array}$
$H A=$ lateral displacement
$\tan 30=\dfrac{H A}{O H}$
$\dfrac{1}{\sqrt{3}}=\dfrac{H A}{8}$
$H A=\dfrac{8}{\sqrt{3}}=4.62 \mathrm{~cm}$
so lateral displacement $=4.62 \mathrm{~cm}$
Hence option $A$ is correct.
Question 9
1 / -0

A bucket of total height 60 cm is half filled with a liquid of refractive index 1.5 and half with another liquid of refractive Index 2. The apparent depth of the bucket for an observer directly above the bucket is

A
45 cm

B
30 cm

C
35 cm

D
40 cm

Solution

Given: A bucket of total height 60 cm is half filled with a liquid of refractive index 1.5 and half with another liquid of refractive Index 2.
To find the apparent depth of the bucket for an observer directly above the bucket
Solution:
As per the given criteria,
Depth of the bucket, $d=60cm$
Refractive index of the liquids, $mu_1=1.5, \mu_2=2$
We know,
Refractive index, $\mu=\dfrac {\text{Real depth}}{\text{Apparent Depth}}$
Bucket is half filled with liquid of refractive index, $\mu_1=1.5$
Therefore, Apparent Depth, $d_1=\dfrac {\dfrac {60}2}{1.5}=20cm$
Other half is filled with liquid of refractive index, $\mu_1=2$
Therefore, Apparent Depth, $d_2=\dfrac {\dfrac {60}2}{2}=15cm$
Total apparent depth,
$D=d_1+d_2\\\implies D=20+15=35cm$
is the required apparent depth of the bucket for an observer directly above the bucket
Question 10
1 / -0

The sky appears blue, due

A
To scattering of light by dust particles

B
To the colour of the sea water

C
To the presence of water in clouds

D
None of these

Solution

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Ray Optics and Optical Test - 46

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Question 1

f=uv/(u+v)f = uv/(u+v)f=uv/(u+v)

v=fu/(u+f)v = fu/(u+f)v=fu/(u+f)

u=fv/(f+v)u = fv/(f+v)u=fv/(f+v)

All of these

Solution

Question 2

1 cm1\ cm1 cm away

1 cm1\ cm1 cm nearer

2 cm2\ cm2 cm away

3 cm3\ cm3 cm away

Solution

Question 3

10cm10cm10cm

15cm15cm15cm

50cm50cm50cm

none of these

Solution

Question 4

3 cm3\ cm3 cm upwards

5 cm5\ cm5 cm upwards

3 cm3\ cm3 cm downwards

5 cm5\ cm5 cm downwards

Solution

Question 5

America

Denmark

Japan

Chile

Solution

Question 6

If n1=n2=ng{ n }_{ 1 }={ n }_{ 2 }={ n }_{ g }n1​=n2​=ng​

If n1=n2{ n }_{ 1 }={ n }_{ 2 }n1​=n2​ and n1<ng{ n }_{ 1 }<{ n }_{ g }n1​<ng​

If n1=n2{ n }_{ 1 }={ n }_{ 2 }n1​=n2​ and n1>ng{ n }_{ 1 }> { n }_{ g }n1​>ng​

under no circumstancesunder\ no\ circumstancesunder no circumstances

Solution

Question 7

f=uv(u+v)f=uv(u+v)f=uv(u+v)

v=fu(u+f)v=fu(u+f)v=fu(u+f)

u=fv(f+v)u=fv(f+v)u=fv(f+v)

None of these

Solution

Question 8

4.62 cm4.62\ cm4.62 cm

5.62 cm5.62\ cm5.62 cm

3.62 cm3.62\ cm3.62 cm

8 cm8\ cm8 cm

Solution

Question 9

45 cm

30 cm

35 cm

40 cm

Solution

Question 10

To scattering of light by dust particles

To the colour of the sea water

To the presence of water in clouds

None of these

Solution

User

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$f = uv/(u+v)$

$v = fu/(u+f)$

$u = fv/(f+v)$

$1\ cm$ away

$1\ cm$ nearer

$2\ cm$ away

$3\ cm$ away

$10cm$

$15cm$

$50cm$

$3\ cm$ upwards

$5\ cm$ upwards

$3\ cm$ downwards

$5\ cm$ downwards

If ${ n }_{ 1 }={ n }_{ 2 }={ n }_{ g }$

If ${ n }_{ 1 }={ n }_{ 2 }$ and ${ n }_{ 1 }<{ n }_{ g }$

If ${ n }_{ 1 }={ n }_{ 2 }$ and ${ n }_{ 1 }> { n }_{ g }$

$under\ no\ circumstances$

$f=uv(u+v)$

$v=fu(u+f)$

$u=fv(f+v)$

$4.62\ cm$

$5.62\ cm$

$3.62\ cm$

$8\ cm$