Question

A parallel beam of monochromatic light of wavelength $5000\stackrel{\circ }{A}$ is incident on a single narrow slit of width $0.001mm$. The light is focused by a convex lens on a screen placed on a focal plane. The first minimum will be formed for the angle of diffraction equal to

• A. $50°$
• B. $15°$
• C. $20°$
• D. $30°$

Answer 1

The correct option is D

$30°$

Step 1: Given data

Wavelength of light, $\lambda =5000A°$

Slit width, $d=0.001mm$

Step 2: To find

The angle of diffraction in which the first minimum will be formed.

Step 3: Formula used

$d\sin \theta =n\lambda$

where $d$ is the slit width, $\lambda$ is the wavelength of light and $n$ is the n^th minima.

Step 4: Calculation for the angle:

Here since the first minimum is considered, $n=1$

$d\sin \theta =n\lambda$

Substituting the values,

$$\begin{gathered} d\sin \theta =n\lambda \\ \Rightarrow \sin \theta =\frac{n\lambda }{d} \\ \Rightarrow \sin \theta =\frac{1\times 5000\times {10}^{-10}}{0.001\times {10}^{-3}} \\ \Rightarrow \sin \theta =\frac{1}{2} \\ \Rightarrow \theta ={\sin }^{-1}\left(\frac{1}{2}\right) \\ \Rightarrow \theta =30° \end{gathered}$$

The first minimum will be formed for the angle of diffraction equal to $30°$

Option D is correct