SATHEE: Optics Question 806

Optics Question 806

Question: A point light source is moving with a constant velocity v inside a transparent thin spherical shell of radius R, which is filled with a transparent liquid. If at t=0 light source is at the center of the sphere, then at what time a thin dark ring will be visible for an observer outside the sphere. The refractive index of liquid with respect to that of shell is $\sqrt{2}$ .

Options:

A) $\frac{R}{\sqrt{2} V}$

B) $\frac{R}{2 V}$

C) $\frac{R}{3 V}$

D) $\frac{R}{\sqrt{3} V}$

Show Answer

Answer:

Correct Answer: A

Solution:

[a] This dark ring will be visible if ray from source gets total internal reflection from the spherical shell.

Let the source at any instant be at point P then at point Q ray will be totally reflected if $θ$ is equal to or greater than critical angle. If QP is equal to x, then $z = \cos θ = \frac{R^{2} + x^{2} - v^{2} t^{2}}{2 R x}$

For $θ$ to be minimum $\frac{d x}{d y} = \frac{2 x (2 R x) - 2 R (R^{2} + x^{2} - v^{2} t^{2})}{4 R^{2} x^{2}} = 0$

$\Rightarrow x = \sqrt{R^{2} - v^{2} t^{2}}$

So, $\cos θ = \frac{2 (R^{2} - v^{2} t^{2})}{2 R \sqrt{R^{2} - v^{2} t^{2}}} = \frac{\sqrt{R^{2} v^{2} t^{2}}}{R}$

For no light come out, $\sin θ \geq \frac{1}{\sqrt{2}} or θ \geq 45^{\circ}$

$\frac{\sqrt{R^{2} - v^{2} t^{2}}}{R} = \frac{1}{\sqrt{2}};$

$t = \frac{R}{\sqrt{2} V}$

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Optics Question 806

Options:

Answer:

Solution:

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एनसीईआरटी व्यायाम वीडियो समाधान

Optics Question 806

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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