Question

In $YDSE$, the spacing between the slits is $d$ and wavelength of light used is $6000\mathring{A}$. If the angular width of a fringes formed on a distant screen is $1^o$, then value of $d$ is:

• A. $1mm$
• B. $0.05mm$
• C. $0.03mm$
• D. $0.01mm$

Answer 1

Answer: (C)

$0.03mm$

Given,

${\theta }_0=1^{\circ }\lambda =6000A^{\circ }=6\times {10}^{-7}m$

For small $\theta$

$sin\theta \approx \theta =\frac{y}{D}$

${\theta }_0=\frac{\lambda D}{d}\times \frac{1}{D}=\frac{\lambda }{d}$

As

${\theta }_0=1^{\circ }=\frac{\pi }{{180}^{\circ }}$

Equating both side

$\frac{\pi }{{180}^{\circ }}=\frac{\lambda }{d}d=\frac{180\times 6\times {10}^{-7}}{\pi }=0.03mm$

Correct option is C.