Question

In a Young's double slit experiment, the slit separation is $1$mm and the screen is $1$m from the slit. For a monochromatic light of wavelength $500$nm, the distance of $3^{rd}$ minima from the central maxima is?

• A. $0.2$ mm
• B. $1.25$ mm
• C. $1.50$ mm
• D. $1.75$ mm

Answer 1

Answer: (A)

$0.2$ mm

We need to obtain 10 maxima of double slit with in central maxima of single slit

For single slit, angular width of central maxima $=\frac{2\lambda }{b}$

For interference pattern

Fringe width$=\frac{\lambda D}{d}$

Therefore $\frac{2\lambda D}{b}=10\frac{\lambda D}{d}$

$b=\frac{d}{2}=0.2mm$