The critical angle of a certain medium is $\sin^{-1}\left(\frac{3}{5}\right)$. The polarizing angle of the medium is
- A$\sin^{-1}\left(\frac{4}{5}\right)$
- B$\tan^{-1}\left(\frac{5}{3}\right)$
- C$\tan^{-1}\left(\frac{3}{4}\right)$
- D$\tan^{-1}\left(\frac{4}{3}\right)$
Explore More
Similar Questions
$A$ beam of light $AO$ is incident on a glass slab $(\mu = 1.54)$ in a direction as shown in the figure. The reflected ray $OB$ is passed through a Nicol prism. On viewing through the rotating Nicol prism,we find that $(\tan 57^{\circ} = 1.54)$.
Difficult
View SolutionWhen a light ray is incident on the surface of a medium,the reflected ray is completely polarized. Then the angle between the reflected and refracted rays is (in $^{\circ}$)
$A$ ray of light is incident at a polarising angle such that its deviation is $24^{\circ}$. Then the angle of incidence is: (in $^{\circ}$)
The angle of polarisation for a medium is $60^o$. What will be the critical angle for this medium?
Medium
View SolutionThe velocity of light in air is $3 \times 10^8 \, ms^{-1}$ and that in water is $2.2 \times 10^8 \, ms^{-1}$. The polarising angle of incidence is.......$^o$
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