Question

In Young’s double slit experiment, double slit of separation 0.1 cm is illuminated by white light. A coloured interference pattern is formed on a screen 100 cm away. If a pin hole is located on this screen at a distance of 2 mm from the central fringe, the wavelength in the visible spectrum which will be absent in the light transmitted through the pin-hole are

Answer 1

571.4 nm, 444.4 nm

Answer 2

The problem involves finding wavelengths that undergo destructive interference at a specific point in a Young's double-slit experiment with white light.

1. Identify the given parameters: slit separation $d$, screen distance $D$, and the position $y$ of the point from the central fringe.
2. Recall the condition for destructive interference: the path difference $\Delta r$ must be $(m + \frac{1}{2})\lambda$, where $m$ is an integer.
3. For YDSE with $y \ll D$, the path difference is $\Delta r = \frac{yd}{D}$.
4. Equate the two expressions: $\frac{yd}{D} = (m + \frac{1}{2})\lambda$.
5. Solve for $\lambda$: $\lambda = \frac{yd}{(m + \frac{1}{2})D}$.
6. Substitute the given values: $d = 10^{-3}$ m, $D = 1$ m, $y = 2 \times 10^{-3}$ m. This gives $\lambda = \frac{2 \times 10^{-6}}{m + 0.5}$ m.
7. Identify the visible spectrum range: 400 nm to 700 nm ($400 \times 10^{-9}$ m to $700 \times 10^{-9}$ m).
8. Set up the inequality: $400 \times 10^{-9} \le \frac{2 \times 10^{-6}}{m + 0.5} \le 700 \times 10^{-9}$.
9. Solve the inequality for the integer $m$: $2.357... \le m \le 4.5$. The possible integer values for $m$ are 3 and 4.
10. Calculate $\lambda$ for $m=3$ and $m=4$: For $m=3$, $\lambda_3 = \frac{2 \times 10^{-6}}{3.5} = \frac{4}{7} \times 10^{-6}$ m $\approx 571.4$ nm. For $m=4$, $\lambda_4 = \frac{2 \times 10^{-6}}{4.5} = \frac{4}{9} \times 10^{-6}$ m $\approx 444.4$ nm.

These are the wavelengths in the visible spectrum that are absent at the given point.