Question

In young's double slit experiment, the slit separation is 1.0 mm and the screen is placed at a distance of 1.0m from the plane of the slits. If the slit separation is changed by 0.05 mm, the fringe width changes by 0.03mm. The wavelength of light used is

• A. 4000A
• B. 5000A
• C. 6000A
• D. 7000A

Answer 1

The correct option is C 6000A

The fringe width $y$, the slit width $d$, the distance between the slit and the screen $D$, and the wavelength $\lambda$ are related by,
$\frac{dy}{D}=\lambda$
$y=\frac{D\lambda }{d}$
Differentiating on both sides, we get
$dy=\frac{-D\lambda }{d^2}dd$
$0.03=\frac{D\lambda }{d^2}0.05$
$\lambda =6000A$