Ray Optics and Optical Test

Result

Ray Optics and Optical Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

A fish looking up through the water sees the outyside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12 cm below the surface, the radius of this circle in cm is

A
$$36\sqrt{7}$$

B
$$36/\sqrt{7}$$

C
$$36/\sqrt{5}$$

D
$$4/\sqrt{5}$$

Solution

The refractive index of water is $$N= \dfrac{4}{3}$$
The critical angle is $$i_c = sin^{-1} \dfrac{1}{n}= sin^{-1} \dfrac{3}{4}$$
if $$r$$ is the radius of the circular horizon and $$h$$ is the distance of the fish from the surface then from the geometry, $$cot \ i_c = \dfrac{r}{h}$$

From trigonometry

$$1+ cot^2 i_c = cosec^2 i_c$$

$$1+ \dfrac{h^2}{r^2} = \dfrac{1}{sin ^2 i_c}$$

$$1+ \dfrac{h^2}{r^2} = \dfrac{4^2}{3^2}$$

$$r= \dfrac{h \sqrt 9}{\sqrt 7} = 12 \times \sqrt{ \dfrac{9}{7}} cm = \dfrac{36}{\sqrt 7} cm$$
Question 2
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A convex lens of focal length f produces a real image 3 times as that size of the object, the distance between the object and the lens is .

A
$$-\left ( \cfrac {2f}3 \right)$$

B
$$-\left ( \cfrac {3f}4 \right)$$

C
$$-\left ( \cfrac {4f}3 \right)$$

D
$$-\left ( \cfrac {3f}2 \right)$$

Solution
Question 3
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The eye can be regarded as a single refracting surface. The radius of this surface is equal to that of cornea (7.8 mm). This surface separates two media of refractive indices 1 and 1.34. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus.

A
2 cm

B
1 cm

C
3.1 cm

D
4.0 cm

Solution

$$\dfrac{1.34}{V} - \dfrac{1}{\infty} = \dfrac{1.34 - 1}{7.8}$$
$$\dfrac{1.34}{v}=\dfrac{34}{780}$$
$$v=\dfrac{1.34\times 780}{34}$$
$$\therefore V = 30.7 mm$$
Question 4
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An object is kept at 0.200 m from a double convex lens or biconvex lens of focal length $$0.0150 m$$ .The position of the image is

A
0.0162m

B
0.600 m

C
8.00 m

D
11.6 m

Solution

$$\frac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$$
$$ \Rightarrow \dfrac{1}{v} + \dfrac{1}{{0.2}} = \dfrac{1}{{0.015}}$$
$$ \Rightarrow \dfrac{1}{v} = \dfrac{{1000}}{{15}} = \dfrac{{10}}{2}$$
$$ \Rightarrow \dfrac{1}{v} = \dfrac{{2000 - 150}}{{30}}$$
$$ \Rightarrow \dfrac{1}{v} = \dfrac{{1850}}{{30}}$$
$$ \Rightarrow v = \dfrac{{30}}{{1850}} = \dfrac{3}{{185}} = 0.0162$$
Hence,
option $$(A)$$ is correct answer.
Question 5
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A ray of light is incident at an angle of $$60^{\circ}$$ on one face of a prism of angle $$30^{\circ}$$. The emergent ray of right makes an angle of $$30^{\circ}$$ with incident ray. The angle made by the emergent ray with second face of prism will be

A
$$0^{\circ}$$

B
$$90^{\circ}$$

C
$$45^{\circ}$$

D
$$30^{\circ}$$
Question 6
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Light enters at an angle of incidence in a transparent rod of refractive index n. For what value of the refractive index of the material of rod the light once entered into it will not leave it through its lateral face whatsoever be the value of angle of incidence?

A
$$n > \surd 2$$

B
$$n = 1$$

C
$$n = 1.1$$

D
$$n = 1.3$$

Solution

From the following figure
$$r+i=90^o \Rightarrow i=90^o -r$$
For ray not to emerge from curved surface $$i > C$$
$$ \Rightarrow \sin i > \sin C \Rightarrow \sin (90^o -r)> \sin C \Rightarrow \cos r > \sin C$$
$$ \Rightarrow \sqrt {1-\sin^2 r} > \dfrac 1n\quad \left\{\because \sin C=\dfrac 1n \right\}$$
$$ \Rightarrow 1-\dfrac {\sin^2 \alpha}{n^2} > \dfrac {1}{n^2} \Rightarrow 1 > \dfrac {1}{n^2} (1+\sin^2 \alpha)$$
$$ \Rightarrow n^2 > 1 \sin^2 \alpha \Rightarrow n > \sqrt 2\quad \left\{\sin i \to 1\right\}$$
$$ \Rightarrow $$ Least value $$=\sqrt 2$$
Question 7
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Find the resolving power of a microscope with cone angle of light falling on the objective equal to $$60^0$$ [Given, $$\lambda = 600 nm$$ and $$\mu_{air} = 1$$]

A
$$1.67 \times 10^6$$

B
$$1.03 \times 10^5$$

C
$$0.67 \times 10^5$$

D
$$8.96 \times 10^6$$

Solution
Question 8
1 / -0

A 2 cm diameter coin rests flat on the bottom of a bowl in which the water is which the water is 20 cm deep $$\left( { \mu }_{ w }=4/3 \right) $$. if the coin is viewed directly from above, what is the apparent diameter:

A
2 cm

B
1.5 cm

C
2.67 cm

D
1.67 cm

Solution

For a plane refracting surface, the lateral magnification is 1.
Thus the image of coin will be of same size as the coin i.e., 2cm
Question 9
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A ray is inclident on boundary separating glass and water Refractive index for glass is $$\frac{3}{2}$$ and refractive index for water is $$\frac{4}{3}$$ critical angle for glass-air boundary is

A
$$sin^{-1}$$$$(\frac{3}{4})$$

B
$$sin^{-1}$$$$(\frac{2}{3})$$

C
$$sin^{-1}$$$$(\frac{8}{9})$$

D
$$sin^{-1}$$$$(\frac{1}{9})$$

Solution

$$\textbf{Hint}$$: Apply Snell's law.
$$\textbf{Step 1}$$:
Given a ray is incident on boundary separating glass and water.
Refractive index for glass is $$\dfrac{3}{2}$$ and refractive index for water is $$\dfrac{4}{3}$$ critical angle for glass-air boundary.
For critical angle, the angle of refraction is $$90^{\circ}$$
$$\textbf{Step 2}$$:
Let us use Snell's law. We know from Snell's law $$\mu_1sin{\theta_c}=\mu_2sin90$$
$$\implies \dfrac{3}{2}sin{\theta_c}=\dfrac{4}{3}\times 1$$
$$\implies sin{\theta_c}=\dfrac{8}{9}$$
$$\implies \theta_c=sin^{-1}\dfrac{8}{9}$$
Thus option C is correct
Question 10
1 / -0

which instrument is used to check whether the body is charged or not?

A
Telescope

B
Lighting conductor

C
Microscope

D
Electroscope

Solution