1.

A prism having retracting angle 5 degree has refractive index 1.5 the angle of minimum deviation is - ​

Answer»

Let δ m be the ANGLE of minimum deviation A=δ m A=δ m μ= sin( 2A )sin 2(A+δ m ) 1.5= sin( 2A )sin( 2A+A ) 1.5= sin( 2A )sin2( 2A ) = sin( 2A )2SIN 2A COS 2A 1.5=2cos 2A 0.75=cos 2A 2A =cos −1 (0.75)2A =41 o A=82 o



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