two slabs with refractive indices 4/3 and 3/2 with thickness of 12cm and 8cm.calculate the effective refractive index....

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We know that the apparent shift in the position of object due to a glass slab of refractive index μ and thickness d is given by
t =d1-1μ
Now according to question,
net shift due to given two  slab is
d11-1μ1+d21-1μ2=d1+d11-d1μ1+d2μ2d1+d2
Let there is a slab of net thickness d1+d2 = d and equivalent refractive index μ, then lateral shift is
t =d1-1μ
sod1-1μ=d1-d1μ1+d2μ2d1+d2 Hence 1μ=d1μ1+d2μ2d1+d2
 Now putting the value of
μ1=4/3μ2=3/2d1=12 cmd1=8 cm 1μ=d1μ1+d2μ2d=1243+83220=1209+163=433×20=4360 Therefore μ =6043 Answer.

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