Ray Optics and Optical Test

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Ray Optics and Optical Test - 49

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Weekly Quiz Competition

Question 1
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An air bubble is inside water. The refractive index of water is $$4/3$$. At what distance from the air bubble should a point object be placed so as to form a real image at the same distance from the bubble:

A
$$2R$$

B
$$3R$$

C
$$4R$$

D
The air bubble cannot form a real image

Solution

$$ \begin{array}{l} { n_{1}=n_{2} } \\ \text { Here, } \\ \text { But here is } \gamma \text { (not possibhe }\\ \text { is this case ) } \\ \text { So the air bubble cannot form a } \\ \text { real inage } \\ \text { Option D is correct. } \end{array} $$
Question 2
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A ray of light is incident on one face of prism with refracting angle $$A(< 90^\circ)$$. The incident ray is normal to the other face of the prism. If C is the critical angle for prism air interface, then the ray will emerge from this faces only if

A
$$cot C < cot A + 1$$

B
$$cot C > cot A + 1$$

C
$$cot A < cot C + 1$$

D
$$cot A > cot C + 1$$

Solution

A ray of light is incident on one face of prism with refracting angle A $$(\angle 90^o)$$ .
The incident ray is normal to each other face of the prism
(Refer to Image)
$$\cfrac {\sin C}{\cos C}=\cfrac {\cfrac {1}{\cos C}}{\sin C}=\cot C$$
and $$\cfrac {\sin A}{\sin B}=\cot A+1$$ .
Question 3
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Two concave lens and a convex of radii 3 cm and 2 cm are joined. What is the refractive index of the combination ?

A
$$\dfrac{3}{5}$$

B
-6

C
$$-\dfrac{3}{5}$$

D
- 12

Solution

It is given that,

$${{R}_{1}}=-3\,cm$$

Focal length of concave lens $${{f}_{1}}=\dfrac{-3}{2}$$

$${{R}_{2}}=2\,cm$$

Focal length of convex lens $${{f}_{2}}=1\,cm$$

Combined focal length is

$$ \dfrac{1}{f}=\dfrac{1}{{{f}_{1}}}+\dfrac{1}{{{f}_{2}}} $$

$$ \dfrac{1}{f}=\dfrac{-2}{3}+1 $$

$$ \dfrac{1}{f}=\dfrac{1}{3}........(1) $$

The relation between redfractive index and focal length is

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

$$ \dfrac{1}{3}=\left( \mu -1 \right)\left[ \dfrac{1}{-3}-\dfrac{1}{2} \right] $$

$$ \dfrac{1}{3}=\left( \mu -1 \right)\left[ \dfrac{-5}{6} \right] $$

$$ \mu =\dfrac{3}{5} $$
Question 4
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An object has an image thrice of its original size when kept at
$$8 cm$$ and $$ 16 cm$$ from a convex lens focal length of the lens is

A
(a)Less than $$ 8 cm $$

B
(b) $$8 cm$$

C
(c)$$ 16 cm$$

D
(d)between $$8$$ and $$16 cm$$

Solution

$$m= \pm 3,$$ using $$m=\dfrac{f}{f+u}$$
For virtual image $$3=\dfrac{f}{f-8} ...(i)$$
For real image $$-3=\dfrac{f}{f-16} ...(ii)$$
Solving $$(i)$$ and $$(ii)$$ we get $$f=12\ cm$$
Question 5
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Three thin prisms are combined as shown in figure. The refractive indices of the crown glass for red, yellow and violet rays are $$\mu_r, $$ $$\mu_y$$ and $$\mu_v$$ respectively and those for the flint glass are $$\mu^{ '}_{r}$$, $$\mu^{'}_{y}$$ and $$\mu^{'}_{v}$$ respectively. Find the $$\dfrac {A'}{A}$$ for which, there is no net angular dispersion.

A
$$\dfrac{\mu_y - 1}{\mu^{'}_{y} - 1}$$

B
$$\dfrac{2\mu_y - 1}{\mu^{'}_{y} - 1}$$

C
$$\dfrac{3\mu_y - 1}{\mu^{'}_{y} - 1}$$

D
$$\dfrac{4\mu_y - 1}{\mu^{'}_{y} - 1}$$

Solution

Total deviation for yellow ray produced

$$\delta y=\delta_{cy}-\delta_{fy}+\delta _{cy}$$

$$=2\delta_{cy}-\delta_{fy}$$

$$=2(\mu_{cy-1})A-(\mu_{-cy}-1)A^1$$

Angular dispersion,

$$\delta_v-\delta_r=[(\mu_{vc}-1)A-(\mu_{vf}-1)A^1+(\mu_{vc}-1)A]$$

$$-[(\mu_{rc}-1)A-(\mu_{rf}-1)A^1+(\mu_{r}-1)A]$$

$$2(\mu_{vc}-1)A-(\mu_{vf}-1)A^1$$

For net angular dispersion to be zero,

$$\delta_r-\delta_r=0$$

$$2(\mu_{vc}-1)A=(\mu_{vf}-1)A^1$$
Question 6
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A light ray is incident a glass sphere at an angle of incidence of $$45^o$$ as shown in figure. Find the total deviation suffered by ray of two successive reflection.

A
$$45^o$$

B
$$30^o$$

C
$$60^o$$

D
$$120^o$$

Solution

Now from a above image
At First reflection

$$1 \times Sin 45^\circ = \sqrt{2} Sin\space r$$

$$r = 30^\circ$$

Since
$$r = i$$ (Both are symmetric)

Show at second reflection

$$\sqrt{2} Sin\space i = 1 \times Sin\space r'$$

$$r' = 45^\circ$$
Now consider quadrilateral A O B C

$$\angle A+\angle O+\angle B+\angle C$$ = 360
$$45+120+45+\angle C = 360$$
$$\angle C = 150$$
Now $$ \angle C +S = 180$$
$$\delta = 30^\circ$$
Hence (B) is correct answer
Question 7
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A ray of light strikes a cubical slab as shown in the figure. Then the geometrical path length traveled by the light in the slab will be:

A
$$2\sqrt {3}$$

B
$$6$$

C
$$\cfrac{2}{\sqrt {3}}$$

D
$$\cfrac{\sqrt{3}}{2}+\cfrac{3}{2}$$

Solution

At the top surface, $$\dfrac{\sin 60}{\sin r}=\sqrt{\dfrac{3}{2}}$$
$$\sin r=\dfrac{1}{\sqrt2}$$
$$r=45^\circ$$
For medium 2,$$45^\circ$$ is greater than its critical angle.
Therefore it undergoes total internal reflection.
From fig. $$AB=BC=CD=2m$$
Total optical path $$=2+2+2=6m$$
Question 8
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A rectangle tank $$6m$$ deep is full of water . By how much does the bottom of the tank appear to be raised when viewed from above. Given $$(\mu_{water} = \frac{4}{3})$$

A
3 m

B
4.5 m

C
1.5 m

D
6 m

Solution
Question 9
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If $$\mu ,$$ represents the refractive index when a light ray goes from medium i to j, then $$_2{\mu _1}{\text{ x}}{{\text{ }}_3}{\mu _2}{\text{ x}}{{\text{ }}_4}{\mu _3}$$ is equal to

A
$$_3{\mu _1}$$

B
$$_3{\mu _2}$$

C
$$\dfrac{1}{{_1{\mu _4}}}$$

D
$$_4{\mu _2}$$

Solution

According to the question,
$$_2\mu_1\times _3\mu_2 \times _4\mu_3 =\dfrac {\mu_1}{\mu_2}\times \dfrac {\mu_2}{\mu_3}\times \dfrac {\mu_3}{\mu_4}=\dfrac {\mu_1}{\mu_4}= _4\mu_1 =\dfrac {1}{_1\mu_4}$$
Question 10
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When a picture drawn on paper is seen through a slab of a transparent materials of thickness $$5cm$$, it appear to be raised by $$1.5 cm$$. Critical angle at the boundary of this transparent material and air is

A
$$\sin^{-1}{\left(\cfrac{2}{3}\right)}$$

B
$$\sin^{-1}{\left(\cfrac{5}{7}\right)}$$

C
$$\sin^{-1}{\left(\cfrac{6}{11}\right)}$$

D
$$\sin^{-1}{\left(\cfrac{7}{10}\right)}$$

Solution

Shift $$=t\left(1-\cfrac{1}{\mu}\right)$$
$$=1.5=5\left(1-\cfrac{1}{\mu}\right)$$
$$0.3=\left(1-\cfrac{1}{\mu}\right)$$
$$\cfrac{1}{\mu}=0.7$$
$$\mu =\cfrac{10}{7}$$
critical angle $$=\sin^{-1}{\cfrac{1}{\mu}}=\sin^{-1}{\left(\cfrac{7}{10}\right)}$$