# 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°.

$^{water}mu&space;_{lens}&space;=&space;frac{mu&space;_{lens}}{mu&space;_{water}}$

= 1.5/1.33

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

$mu&space;=&space;frac{Sinfrac{A+delta&space;_{m}}{2}}{Sinfrac{A}{2}}$

$9/8&space;=&space;frac{sinfrac{60+delta&space;_{m}}{2}}{Sin&space;30}$

$9/16&space;=&space;[Sin&space;frac{60&space;+&space;delta&space;_{m}}{2}]$

$0.56&space;=&space;[Sin&space;frac{60&space;+&space;delta&space;_{m}}{2}]$

$34.3^{0}&space;=&space;frac{60&space;+&space;delta&space;_{m}}{2}$

$68.6^{0}&space;=&space;60&space;+&space;delta_{m}$

We get,

$delta&space;_{m}&space;=&space;8.6^{0}$

Therefore, The angle of minimum deviation is 8.60