Two equilateral prisms made of material of refractive indices 1.50 and 1.68 are placed one after another with their refracting edge furned towards the same side as shown in figure. If both the prisms are set for minimum deviation, find (i) the angle between the bases of the two prisms and (ii) the focal deviation of a parallel beam of light incident on the combination.

Explanation

In the minimum deviation position, `r = (A)/(2) = (60^(@))/(2)` for both the prisms. Since `A = 60^(@)` for both the prisms. For the first prism, `mu_(1) = 1.50` and `sin i_(1) = mu, sin r_(1) = 1.50, sin 30^(@) = 0.75`

`:. i_(1) = 48^(@)35'`
For the second prism, `mu_(2) = 1.68` and `sin i_(1) = mu_(2)sin r_(1)' = 1.68 sin 30^(@) = 0.84`
`i'_(1) = 57^(@)9'`
Let the angle between the bases of the prism be `alpha`.
`alpha = 360^(@) - (60^(@) + beta + 60^(@))`
But `beta + 90^(@) - i_(1) + 90^(@) - i_(1) = 180^(@)`
`beta = i_(1) + i'_(1) = 48^(@)35^(@) + 57^(@)9' = 105^(@)44'`
`:. alpha = 360^(@) - (120^(@) + 105^(@)44') = 134^(@)16'`
(ii) In the minimum deviation position, `i = (A+delta)/(2)`
`delta = 2i - A`
For the first prism, `delta_(1) = 2i_(1) - A = 2(48^(@)35') - 60^(@) = 37^(@)10'`
For the second prism, `delta_(2) = 2i_(1) - A = 2(57^(@)9') - 60^(@) = 54^(@)18'`