In an equilateral prism of `mu=1.5,` the condition for minimum deviation is fulfilled. If face AC is polished
Find the net deviation.
If the system is placed in water what will be the net deviation for same angle of incidence? Refractive index of water `=4/3.`

Answer

A, B

Explanation

At minimum deviation,
`r_1=r_2=A/2=30^@`
Applying `mu=(sin i_1)/(sin r_1) or 1.5=(sin i_1)/(sin 30^@)`
We get `i_1=48.6^@`

`delta_(Total)=delta_P+delta_Q+delta_R`
`=(48.6^@-30^@)+(180^@-2xx30^@)+(48.6^@-30^@)`
`=157.2^@`
`4/3sin(48.6^@)=1.5sinr_1`
Solving we get, `r_1=41.8^@`
`r_2=60^@-r_1=18.2^@`
`r_3=60^@-r_2=41.8^@`
`:. i_3=48.6^@`
Hence,
`delta_(Total)=(48.6^@-41.8^@)+(180^@-2xx18.2^@)+(48.6^@-41.8^@)`
`=157.2^@`