A prism can produce a minimum deviation d in a light beam. If three such prisms are combined, the minimum deviation that can be produced in this beam is:

  • Option 1)

    0

  • Option 2)

    d

  • Option 3)

    2d

  • Option 4)

    3d

 

Answers (1)

 

Condition of maximum deviation -

i=e

 

 

- wherein

\delta _{min}= 2i-A

\mu = \frac{\sin \left ( \frac{\delta _{m}+A}{2} \right )}{\sin \left ( A/2 \right )}

 

  value of a minimum deviation

d_{net} = d-d+d =d


Option 1)

0

Incorrect

Option 2)

d

Correct

Option 3)

2d

Incorrect

Option 4)

3d

Incorrect

Preparation Products

Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout BITSAT-JEE Main 2021

An exhaustive E-learning program for the complete preparation of JEE Main and Bitsat.

₹ 27999/- ₹ 16999/-
Buy Now
Exams
Articles
Questions