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Physics Question on Optics

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 cm away. The wavelength of light used is _______.
Fill in the blank with the correct answer from the options given below

A

2.5×10−7 m2.5 \times 10^{-7} \ \text{m}

B

4×10−7 m4 \times 10^{-7} \ \text{m}

C

5×10−7 m5 \times 10^{-7} \ \text{m}

D

7.5×10−7 m7.5 \times 10^{-7} \ \text{m}

Answer

5×10−7 m5 \times 10^{-7} \ \text{m}

Explanation

Solution

The width of the central maximum in single-slit diffraction is given by:
w=2λDaw = \frac{2 \lambda D}{a}
where,
w=5 mm=5×10−3 mw = 5 \ \text{mm} = 5 \times 10^{-3} \ \text{m} (width of central maximum),
D=50 cm=0.5 mD = 50 \ \text{cm} = 0.5 \ \text{m} (distance to the screen),
a=0.1 mm=1×10−4 ma = 0.1 \ \text{mm} = 1 \times 10^{-4} \ \text{m} (slit width),
λλ is the wavelength of light.
Rearranging the formula forλ λ:
λ=wa2D=5×10−3×1×10−42×0.5=5×10−7 m\lambda = \frac{wa}{2D} = \frac{5 \times 10^{-3} \times 1 \times 10^{-4}}{2 \times 0.5} = 5 \times 10^{-7} \ \text{m}