Question

Light of wave length 6.5 $\times {10}^{-7}$ m is made incident on two slits 1 mm apart. The distance between third dark fringe and fifth bright fringe on a screen distant 1 m from slits will be

• A. 0.325 mm
• B. 0.65 mm
• C. 1.625 mm
• D. 3.25 mm

Answer 1

Answer: (C)

1.625 mm

x${}_5$ = n$\lambda \frac{D}{d}=\frac{5\times 6.5\times {10}^{-7}\times 1}{{10}^{-3}}$ = 32.5 $\times$ 10${}^{-4}$ m
x${}_3$ = (2n - 1) $\frac{1}{2}\frac{D\lambda }{d}=\frac{5\times 6.5\times {10}^{-7}\times 1}{2\times {10}^{-3}}$ = 16.25 $\times$ 10${}^{-4}$ m
x${}_5$ - x${}_3$ = 16.25 $\times$ 10${}^{-4}$m = 1.625 mm.