Two slabs of same thickness 1 cm each are placed as shown.
Find the refractive index of a slab of thickness 3 cm which will produce the same normal shift in position of object (O), as produced by combination of slabs.
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Solution
Let μeq be the R.I of slab of thickness t=3cm ⇒Snet=S1+S2
or t(1−1μeq)=t1(1−1μ1)+t2(1−1μ2) 3(1−1μeq)=1(1−23)+1(1−12) ⇒3(1−1μeq)=13+12=56
or, 1−1μeq=518
or, 1μeq=1−518=1318 ∴μeq=1813=1.38
Why this question?
Caution: The formula μeq=2μ1μ2μ1μ2 is only applicable when sum of thickness of two slabs is equal to thickness of equivalent slab.