$A$ ray of light passes from a medium $A$ having refractive index $1.6$ to the medium $B$ having refractive index $1.5$. The value of the critical angle of medium $A$ is . . . . . . .
- A$\sin^{-1} \sqrt{\frac{16}{15}}$
- B$\sin^{-1} \left(\frac{16}{15}\right)$
- C$\sin^{-1} \left(\frac{1}{2}\right)$
- D$\sin^{-1} \left(\frac{15}{16}\right)$
Explore More
Similar Questions
On which principle does an optical fiber work?
Easy
View SolutionThe critical angle for total internal reflection of light going from medium $I$ to medium $II$ is given by the relation $\tan i_C = \frac{5}{9}$. The refractive index of medium $I$ with respect to medium $II$ is:
Medium
View SolutionFor the given incident ray as shown in the figure,what is the minimum refractive index of the prism required for total internal reflection of this ray?
Difficult
View SolutionThe refractive index of glass is $\frac{3}{2}$ and that of water is $\frac{4}{3}$. The critical angle for a ray of light going from glass to water is
$A$ fish looking from within water sees the outside world through a circular horizon. If the fish is $\sqrt{7} \ cm$ below the surface of water,what will be the radius of the circular horizon in $cm$?
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo