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Width of the principal maximum on a scre...

Width of the principal maximum on a screen at a distance of 50 cm from the slit having width 0.02 cm is `312.5xx10^(-3)cm`. If waves were incident normally on the slit, then wavelength of the light from the source will be

A

`6000Å`

B

`6250Å`

C

`6400Å`

D

`6525Å`

Text Solution

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The correct Answer is:
To solve the problem, we need to determine the wavelength of light based on the given parameters of the slit and the width of the principal maximum on the screen. ### Step-by-Step Solution: 1. **Identify Given Values**: - Width of the slit (A) = 0.02 cm - Distance from the slit to the screen (D) = 50 cm - Width of the principal maximum (W) = 312.5 x 10^(-3) cm 2. **Calculate the Position of the First Minimum**: - The width of the principal maximum is the distance between the first minima on either side of the central maximum. Therefore, the distance from the center to the first minimum (y1) is: \[ y_1 = \frac{W}{2} = \frac{312.5 \times 10^{-3}}{2} = 156.25 \times 10^{-3} \text{ cm} \] 3. **Use the Condition for Minima**: - The condition for the first minima in single-slit diffraction is given by: \[ A \sin \theta = n \lambda \] - For the first minimum, \( n = 1 \), so: \[ A \sin \theta = \lambda \] 4. **Approximate \(\sin \theta\)**: - Since the angle \(\theta\) is small, we can use the small angle approximation: \[ \sin \theta \approx \tan \theta \approx \frac{y_1}{D} \] 5. **Substituting Values**: - Substitute \( \sin \theta \) in the equation: \[ A \left(\frac{y_1}{D}\right) = \lambda \] - Plugging in the values: \[ 0.02 \times \left(\frac{156.25 \times 10^{-3}}{50}\right) = \lambda \] 6. **Calculate \(\lambda\)**: - Perform the calculation: \[ \lambda = 0.02 \times \frac{156.25 \times 10^{-3}}{50} \] \[ = 0.02 \times 3.125 \times 10^{-3} = 6.25 \times 10^{-5} \text{ cm} \] 7. **Convert to Angstroms**: - To convert centimeters to angstroms (1 cm = 10^8 angstroms): \[ \lambda = 6.25 \times 10^{-5} \text{ cm} \times 10^8 \text{ angstroms/cm} = 6250 \text{ angstroms} \] ### Final Answer: The wavelength of the light from the source is \( \lambda = 6250 \text{ angstroms} \). ---
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