Ray Optics and Optical Test

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Ray Optics and Optical Test - 46

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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
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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