(a) will be same

(b)will be 2 mm

(c)will be 4 mm

(d)will be 1mm

Since $10th$ maxima is 3 cm from from central maxima the fringe width $\beta =\large\frac{3}{10}$

$\qquad= 0.3 cm$

$\qquad= 3 mm$

Now when YDSE is immersed in a liquid of refractive index $\mu $ new fring width $\beta'=\large\frac{\beta}{\mu}$

$\qquad= \large\frac{0.3}{1.5} $$=0.2 \;cm$

$\qquad= 2 \;mm$

Hence b is the correct answer

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