Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
577 views
in Physics by (30.0k points)
closed by

Show that for a material with refractive index μ ≥ √2, light incident at any angle shall be guided along a length perpendicular to the incident face.

1 Answer

+1 vote
by (26.9k points)
selected by
 
Best answer

Any ray entering at an angle shall be guided along AC if the angle the ray makes with the face AC(ϕ) is greater than the critical angle.

\(⇒sin\,r≥\frac{1}{μ}\)

\(⇒cos\,r≥\frac{1}{μ}\)

Or, 1 – cos2 r ≤ 1 − \(\frac{1}{μ^2}\)

i.e. sin2 r ≤ 1 − \(\frac{1}{μ^2}\)

Since sin i = μ sin r

\(\frac{1}{μ^2}sin^2\,i\) ≤ \(1-\frac{1}{μ^2}\)

Or, sin2 i ≤ μ2 − 1

The smallest angle ϕ shall be when \(i=\frac{π}{2}\). If that is greater than the critical angle then all other angle of incidence shall be more than the critical angle.

Thus 1 ≤ μ2 − 1

Or, μ2 ≥ 2

⇒ μ ≥ √2

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...