A Polarised and an analyser are oriented so that the maximum light is transmitted. what is the fraction of maximum light transmitted when analyser is rotated through (1) 30 degree (2) 60 degree ?

Dear Harshita
since
I=I0/2(when light pas through polaraiser)
and
I'=I cos2theta
so
(i)
finally we have
I'=(I/2)cos²30°   
=(I/2)(3/4)
=(3/8)
ii)
I'= (I/2) cos² 60°
= (I/2) (1/4)
= 1/8
Regards
Arun (askIITians forum expert)