Using light from a monochromatic source to study diffraction in a single slit of width 0.1 mm, the linear width of central maxima is measured to be 5 mm on a screen held 50 cmaway. The wavelength of light used is _______.

The correct option is 5 × 10⁻⁷ m.

Step-by-Step Derivation and Explanation:

In single-slit diffraction, the angular position of the first minimum on either side of the central maximum is given by the condition:

$$d\sin \left(\theta \right)=\lambda$$
where:
• $d$ is the slit width,
• $\theta$ is the angle of diffraction for the first minimum,
• $\lambda$ is the wavelength of the light used.

For small angles of diffraction, we can approximate $\sin \left(\theta \right)\approx \theta$. Therefore, the angular half-width of the central maximum is:

$$\theta =\frac{\lambda }{d}$$
The linear half-width on a screen placed at a distance $D$ is $x=D\theta$. Thus, the total linear width of the central maximum ($W$) which extends from the first minimum on one side to the first minimum on the other side is given by:

$$W=2x=\frac{2\lambda D}{d}$$

1. Identify the given values from the problem:
• Width of the slit, $d=0.1\text{mm}={10}^{-4}\text{m}$
• Distance of screen from the slit, $D=50\text{cm}=0.5\text{m}$
• Linear width of the central maximum, $W=5\text{mm}=5\times {10}^{-3}\text{m}$

2. Rearrange the formula to solve for the wavelength ($\lambda$):

$$\lambda =\frac{W·d}{2D}$$

3. Substitute the values into the rearranged equation:

$$\lambda =\frac{(5\times {10}^{-3}\text{m})\times \left({10}^{-4}\text{m}\right)}{2\times 0.5\text{m}}$$
$$\lambda =\frac{5\times {10}^{-7}}{1.0}\text{m}$$
$$\lambda =5\times {10}^{-7}\text{m}$$

Hence, the wavelength of the light used is indeed 5 × 10⁻⁷ m.