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anonymous
 4 years ago
The wavelength of light in two liquids "x" and "y" is 3500 angstroms and 7000 angstroms respectively. Find the critical angle of "x" relative to "y".
anonymous
 4 years ago
The wavelength of light in two liquids "x" and "y" is 3500 angstroms and 7000 angstroms respectively. Find the critical angle of "x" relative to "y".

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apoorvk
 4 years ago
Best ResponseYou've already chosen the best response.0You know the wavelengths of the red and violet, you can find relative refractive indices for two lights (by assuming yellow light as a constant). Then you can use the formula for critical angle, since you already have the refractive indices. (just calculate everything in terms of an assumed constant 'x', and in the end you'll find that the 'x' gets cancelled)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can you expalin the steps please?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Ishaan94 please try.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You need to tell me what's critical angle?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The angle of incidence for which the angle of refraction is 90 degrees is called "critical angle".

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh hmm, so relative reflective index must be \(\large n = \frac{\lambda_x}{\lambda_y}\), right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1334817280641:dw\[n = \frac{\sin\theta_i}{\sin \theta_r}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(\theta_r \to 90 \implies \sin\theta_r \to1\)\[\implies n = \sin\theta_i\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wow. Nice. Thankyou Aadarsh for posting it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah. Got it. So, it is: \[\sin 30 = 3500/7000 = 1/2 \] So, answer = sin 30 degrees, rite?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks Bhaiiiiiiiiijan!!!
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