A glass prism of refracting angle 60° and refractive index 1.5 is completely immersed in water of refractive index 1.33. Calculate the angle of minimum deviation of the prism in this situation, sin-1 0.56 = 34.3°.

\(\begin{array}{l}^{Water}\mu _{lens}=\frac{\mu _{lens}}{\mu _{water}}\end{array} \)

= 1.5/1.33

= (3/2)/(4/3) = 9/8

\(\begin{array}{l}\mu =\frac{Sin\frac{A+\delta _{m}}{2}}{Sin\frac{A}{2}}\\\frac{9}{8}=\frac{Sin\frac{60+\delta _{m}}{2}}{Sin 30}\\\frac{9}{16}= \left [ Sin\frac{60+\delta _{m}}{2} \right ]\\0.56 = \left [ Sin\frac{60+\delta _{m}}{2} \right ]\\34.3^{0}= \frac{60+\delta _{m}}{2}\\68.6^{0}= 60+\delta _{m}\end{array} \)

We get,

\(\begin{array}{l}\delta _{m} = 8.6^{0}\end{array} \)

Therefore, The angle of minimum deviation is 8.60

Was this answer helpful?

 
   

3.5 (2)

(8)
(3)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published.

*

*

close
close

DOWNLOAD

App Now

Ask
Question