In a Fraunhofer diffraction of a single slit, when a slit is illuminated by a light of wavelength $6480 \text{Å}$, angular width of central maximum is measured. When the slit is illuminated by light of another wavelength '$λ$' the angular width decreases by $25\%$. The value of $λ$ in $\text{Å}$ units is

Question

In a single slit diffraction pattern, a light of wavelength $ 6000 \, \text{Å} $ is used. The distance between the first and third minima in the diffraction pattern is found to be 3 mm when the screen is placed 50 cm away from the slits. The width of the slit is ______ $ \times 10^{-4} $ m.

Answer 1

Step 1: Set up the proportion:
$\frac{\lambda'}{\lambda} = \frac{\text{new width}}{\text{old width}} = \frac{75}{100}$.

Step 2: Calculate:
$\lambda' = \frac34 \times 6480 = 4860\ \text{ angstrom}$.

Final Answer:
$\lambda = 4860\ \text{ angstrom}$, option (C). \[ \boxed{4860\ \text{ angstrom}} \]

Show Hint

The Correct Option is C

Solution and Explanation

Top Questions on single slit diffraction

Questions Asked in MHT CET exam