A light of wavelength $$6000\,\mathop {\text{A}}\limits^ \circ $$ shines on two narrow slits separated by a distance $$1.0\,mm$$ and illuminates a screen at a distance $$1.5\,m$$ away. When one slit is covered by a thin glass plate of refractive index 1.8 and other slit by a thin glass plate of refractive index $$\mu ,$$ the central maxima shifts by $$0.1\,rad.$$ Both plates have the same thickness of $$0.5\,mm.$$ The value of refractive index $$\mu $$ of the glass is
A.
1.4
B.
1.5
C.
1.6
D.
None of these
Answer :
1.6
Solution :
Change in path difference for any point on screen is $$\left| {\mu - 1.8} \right|t.$$
For central maxima, phase difference $$= 0.$$
Hence,
$$\eqalign{
& d\sin \theta - \left| {\mu - 1.8} \right|t = 0 \cr
& \Rightarrow d\theta = \left| {\mu - 1.8} \right|t\,\,\left[ {q\,{\text{is very small and is in radian}}} \right] \cr
& \Rightarrow \left| {\mu - 1.8} \right| = \frac{{{{10}^{ - 3}} \times 0.1}}{{0.5 \times {{10}^{ - 3}}}} = 0.2 \cr
& \Rightarrow \mu = 2\,\,{\text{or}}\,\,1.6 \cr} $$
Releted MCQ Question on Optics and Wave >> Wave Optics
Releted Question 1
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