Question

for what vlaue of critical angle velocity of light is 1.5×10^8 m/s in that medium
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Answer 1

Answer:

A ray travels from medium M1 to the medium M2 with an angle of incidence theta . The ray suffers total internal reflection. Then the value of the angle of incidence ...

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Answer 2

For total internal reflection to occur, the angle of incidence (θ) must be greater than critical angle $\large\rm { \theta_{c}}$

$\large\rm { \theta > \sin^{-1} \bigg ( \frac{n_{r}}{n_{d}} \bigg ) }$

We know that, $\Large\rm { \frac{n_{r}}{n_{d}} = \frac{v_{1}}{v_{2}} }$

so, $\large\rm { \theta > \sin^{-1} \bigg ( \frac{v_{1}}{v_{2}} \bigg )}$

$\large\rm { \therefore \theta > \sin^{-1} \bigg ( \frac{1.5 \times 10^{8}}{2 \times 10^{8}} \bigg ) }$

$\large\rm { \implies \theta > sin^{-1} \bigg ( \frac{3}{4} \bigg )}$