Ray Optics and Optical Test

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Ray Optics and Optical Test - 52

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Question 1
1 / -0

In refraction of light through a glass slab, the directions of the incident ray and refracted ray are:

A
perpendicular to each other

B
non-parallel to each other

C
parallel to each other

D
intersecting each other

Solution

R.E.F image
The gap t is called
"lateral Displacement"
$$ t = \frac{d sin (i-r_{1})}{cosr_{2}} $$
Question 2
1 / -0

The focal point of an equi- convex lens whose refractive index is $$1.5$$ is $$10\ cm$$ in air. Its focal point inside a liquid of refractive index $$1.25$$ is

A
$$20\ cm$$

B
$$25\ cm$$

C
$$30\ cm$$

D
$$50\ cm$$

Solution

Given that,

Focal length $$f=10\,cm$$

Refractive index $${{\mu }_{1}}=1.5$$

Refractive index $${{\mu }_{2}}=1.25$$

We know that,

$$ \dfrac{1}{f}=\left( n-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right) $$

$$ \dfrac{1}{10}=\left( 1.5-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)....(I) $$

$$ \dfrac{1}{f}=\left( 1.25-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)....(II) $$

Now, divided equation (I) by equation (II)

$$ \dfrac{f}{10}=\dfrac{0.5}{0.25} $$

$$ f=20\,cm $$

Hence, the focal length is $$20\ cm$$
Question 3
1 / -0

An object placed at a distance of 16 cm from first principal focus of convex lens, produces a real image at a distance of 36 cm from its second principal focus. Then the focal length of the lens is :

A
$$9 cm$$

B
$$25 cm$$

C
$$24 cm$$

D
$$17 cm$$

Solution

Let focal length of the lens be $$f$$,

Given,

Object distance $$=16+f=u$$

Image distance $$=36+f=v$$

Distance of 1st and 2nd principal focus is same from the optical centre

From, lens formula

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

$$\cfrac{1}{f}=\cfrac{1}{36+f}+\cfrac{1}{16+f}$$

$$\cfrac{1}{f}=\cfrac{16+f+36+f}{(36+f)(16+f)}$$

On solving

$$f=24cm$$
Question 4
1 / -0

In a thin spherical fish bowl of radius $$10\ cm$$ filled with water of refractive index $$\dfrac{4}{3}$$, there is a fish at a distance of $$4\ cm$$ there is a small fish at a distance of $$4\ cm$$ form the center $$C$$ as shown. The image of the fish appears, when seen from the point $$D$$ would be

A
$$5.2\ cm$$

B
$$4.8\ cm$$

C
$$2.6\ cm$$

D
$$7.4\ cm$$

Solution

$$v \to image\,position$$
$$u \to object\,position$$
$$R \to Radius\,of\,surface$$
$$\dfrac{{{n_2}}}{v} - \dfrac{{{n_1}}}{u} = \dfrac{{{n_2} - {n_1}}}{R}$$
$$ \Rightarrow \dfrac{1}{v} - \dfrac{{\left( {4/3} \right)}}{{\left( { - 6} \right)}} = \dfrac{{1 - 4/3}}{{\left( { - 10} \right)}}$$
$$ \Rightarrow \dfrac{1}{v} + \dfrac{4}{{3 \times 6}} = \dfrac{1}{{3 \times 10}}$$
$$ \Rightarrow \dfrac{1}{v} = \dfrac{1}{{30}} - \dfrac{4}{{18}}$$
$$ \Rightarrow \dfrac{1}{v} = \dfrac{{ - 17}}{{90}}$$
$$ \Rightarrow v = \dfrac{{ - 90}}{{17}} = 5.2\,cm$$
Hence,
option $$(A)$$ is correct answer.
Question 5
1 / -0

A plane mirror is made of glass slab $$(n=1.5)2.5 cm$$ thick and silvered back. A point object placed $$5 cm$$ in front the unsilvered face of the mirror. The position of final image is:

A
$$12 cm$$ from unsilured face

B
$$14.6 cm$$ from unsilvered face

C
$$5.67 cm$$ from unsilvered face

D
$$8.33 cm$$ from from unsilvered face

Solution

$$ \begin{array}{l} \text { Let } I_{1}, I_{2} \text { and } I_{3} \text { be the image } \\ \text { formed by } \end{array} $$
$$ \begin{array}{l} \text { (i) refraction from } A D C \\ \text { (ii) reflection from BEF. } \\ \text { (iii) Again refraction from ADC. } \\ \text { Then } B I_{1}=5 n=5(1.5)=7.5 \\ =7.5 \mathrm{~cm} \\ \text { Now } \epsilon I_{1}=(7.5+2.5) \mathrm{cm}=10 \mathrm{~cm} \\ \in I_{2}=10 \mathrm{~cm} \text { behind the mirror. } \end{array} $$
$$ \begin{array}{l} \text { Now, } B I_{2}=10+2.5=12.5 \mathrm{~cm} \\ \therefore B I_{3}=\frac{12.5}{\pi}=\frac{12.5}{1.5} \\ =8.33 \mathrm{~cm} \\ \text { Hence, } D \text { is the correct answer } \end{array} $$
Question 6
1 / -0

A student while observing a coin placed at the bottom of the container filled with water finds it to be at 2 m depth .To observe the coin at its actual depth he should use

A
a convex lens

B
a concave lens

C
a plane mirror

D
a concave mirror

Solution

The ray coming from coin get diverted at the surface of water$$.$$
$$\therefore $$ He should use a converging lens to get the actual depth$$.$$
Hence$$,$$ option $$(A)$$ is correct$$.$$
Question 7
1 / -0

How much water should be filled in a container $$21 cm$$ in height, so that it appears half filled when viewed from the top of the container? (given that $$a \mu_\omega$$ = 4/3)

A
$$8.0 cm$$

B
$$10.5 cm$$

C
$$12.0 cm$$

D
None of the above

Solution

To see the container half-filled from top, water should be filled up to height $$x$$ so that bottom of the container should appear to be raised upto height $$(21-x)$$
As shown in figure appearent depth $$h'=(21-x)$$
Real depth $$h=x$$
$$\therefore \mu =\dfrac {h}{h'}\Rightarrow \dfrac {4}{3}=\dfrac {Bottom}{21-x}\Rightarrow x=12\ cm$$
Question 8
1 / -0

A ray of light passes normally through a slab $$(\mu = 1.5)$$ of thickness $$'t'$$. If the speed of light in vacuum be $$'c'$$ then time taken by the ray to go across the slap will be

A
$$\dfrac { t }{ c } $$

B
$$\dfrac { 3t }{ 2c } $$

C
$$\dfrac { 2t }{ 3c } $$

D
$$\dfrac { 4t }{ 9c } $$

Solution

$$\begin{array}{l} \eta =\dfrac { c }{ v } \\ \Rightarrow v=\dfrac { c }{ { 1.5 } } =\dfrac { { 2c } }{ 3 } \\ time=\dfrac { { 3t } }{ { 2c } } \\ Hence, \\ option\, \, B\, \, is\, correct\, \, answer. \end{array}$$
Question 9
1 / -0

Choose the wrong statement related to refraction of light

A
Twinkling of stars

B
Oval shaped of sun in morning and evening

C
Object in water appears bigger in size

D
Red light undergoes dispersion, while passing through prism

Solution

* Red light undergoes dispersion, while passing thruugh brism the bhemenon is called dispersion 8 if leads to it is due to the long-wavelength of red light that is bents the least while violet bends the most , that's why red gets segrigated out, its not due to refraction of light. While 'A'B' C' option's are due to refraction of light
Hence abtion $$D$$ is correct.
Question 10
1 / -0

Four identical lenses are kept one beside the other on the same optical as shown in the figure. The right surface of rightmost lens is silvered. Focal length of lens is 20 cm and radius of silvered is 20 cm. The focal length of the combined system is

A
2 cm

B
-2 cm

C
5 cm

D
-5 cm

Solution

$$\begin{array}{l}\text { for net focus of lenses } a, b, c \& d \text { we use formula } \\\frac{1}{F_{n e t}}=\frac{1}{f_{a}}+\frac{1}{f_{b}}+\frac{1}{f_{c}}+\frac{1}{f_{d}}=\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}=\frac{4}{20}=\frac{1}{5}&={F_{net}}=5cm\end{array}$$
for net focus of the systern we use formula
$$\frac{1}{F_{\text {net }}}=\frac{-2}{f_{\text {lens }}}+\frac{a_{2}}{R_{\text {mirror }}}=\frac{-2}{5}+\frac{2}{(-20)}$$
$$F_{\text {net }}=-2 \mathrm{~cm}$$
Hence $$B^{\prime}$$ is correct