1.

Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism of given prism. Then the angle of prism is ...... (sin 48^@36. = 0.75)

Answer»

`41^@24.`
`60^@`
`80^@`
`82^@48.`

Solution :`eta=(SIN((A+delta_m)/(2)))/(sin""A/2)=(sinA)/(sin""A/2)[because delta_m=A]`
`1.5=(2sin""A/2os""A/2)/(sin""A/2)`
`0.75=cos""A/2`
`0.75=sin(90^@-A/2)`
`therefore 48^@36.=90^@-A/2`
`therefore A/2=90^@-48^@36.`
`therefore A/2=41^@24.`
`therefore A=82^@48.`


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