Question

In Young's double slit experiment, the spacing between the slits is \(d\) and wavelength of light used is \(6000\,\text{\AA}\). If the angular width of a fringe formed on a distance screen is \(1^\circ\), the value of \(d\) is

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

Hint:

Angular fringe width in YDSE is directly \(\theta = \dfrac{\lambda}{d}\). Convert degrees to radians before substituting.

Answer 1

The Correct Option is C

Step 1: Use angular fringe width formula.
Angular width of fringe is given by:
\[ \theta = \frac{\lambda}{d} \] Step 2: Convert given data to SI units.
\[ \lambda = 6000\,\text{\AA} = 6000 \times 10^{-10} = 6\times 10^{-7}\,m \] \[ \theta = 1^\circ = \frac{\pi}{180} \approx 0.01745\ \text{rad} \] Step 3: Find slit separation \(d\).
\[ d = \frac{\lambda}{\theta} = \frac{6\times 10^{-7}}{0.01745} \approx 3.44\times 10^{-5}\,m \] Step 4: Convert into mm.
\[ 3.44\times 10^{-5}\,m = 3.44\times 10^{-2}\,mm \approx 0.03\,mm \] Final Answer: \[ \boxed{0.03\ \text{mm}} \]