With the one slit 25mm from the biprism and the other 16 mm and using light of wavelength 1200 nm, fringes were formed on a screen 0.5 m from the single slit.The fringe width is

0.03 cm

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0.03 mm

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0.3 cm

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0.3 m

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Solution

The correct option is B 0.03 mm
$d = \sqrt{d_{1} d_{2}} = 20 m m$
Fringe width= $\frac{D λ}{d}$
$= \frac{0.5 \times 1200 \times 10^{- 9}}{20 \times 10^{- 3}} m = 0.03 m m$

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With the one slit 25mm from the biprism and the other 16 mm and using light of wavelength 1200 nm, fringes were formed on a screen 0.5 m from the single slit.The fringe width is

The correct option is B 0.03 mmd=√d1d2=20mmFringe width=Dλd=0.5×1200×10−920×10−3m=0.03mm

The correct option is B 0.03 mm
$d = \sqrt{d_{1} d_{2}} = 20 m m$
Fringe width= $\frac{D λ}{d}$
$= \frac{0.5 \times 1200 \times 10^{- 9}}{20 \times 10^{- 3}} m = 0.03 m m$