Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
The refractive index (`mu`) of a prism of angle A and minimum deviation, `delta_(m)` is given by
`mu=sin((4+delta_(m))/2)/sin(A//2)`
Given, `mu=cos””A/2`
`therefore cot””A/2=sin((A+delta_(m))/2)/sin(A//2)`
`(cosA//2)/(sinA//2)=(sin””((A+delta))/2)/sin(A//2)rArrcos””A/2=sin((A+delta_(m))/2)`
`therefore sin(90^(@)-A/2)=sin((A+delta_(m))/2)`
`rArr90^(@)-A/2=(A+delta_(m))/2rArr180^(@)-A=A+delta_(m)`
`rArr` Angle of minimum deviation, `delta_(m)=180^(@)-2A=pi-2A`
Correct Answer – B
(b) Refractive index,`mu=sin((A+D_(m))/2)/(sin””A/2)`
`rArrcot””A/2=sin((A+D_(m))/2)/(sin””A/2)`
`rArr(cot””A/2)/(sin””A/2)=sin((A+D_(m))/2)/(sin””A/2)`
`sin((pi)/2-A/2)=sin((A+D_(m))/2)`
`pi/2-A/2=A/2+(D_(m))/2`
`rArrD_(m)=pi-2A`
Angle of minimum deviation, `D_(m)=180^(@)-2A`