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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
X-rays

Concepts Of Physics MCQ Edition [Volume 2] · Physics · X-rays

Consider a monochromatic X-ray beam with a wavelength of 100 pm...

Consider a monochromatic X-ray beam with a wavelength of 100 pm100 \text{ pm} passing through a Young's double slit. The resulting interference pattern is recorded on a photographic plate located 40 cm40 \text{ cm} from the slit. Determine the required separation between the slits if the successive maxima on the screen are to be separated by a distance of 0.1 mm0.1 \text{ mm}.

Options

  1. A4×10−8 m4 \times 10^{-8} \text{ m}
  2. B2.5×10−7 m2.5 \times 10^{-7} \text{ m}
  3. C4×10−6 m4 \times 10^{-6} \text{ m}
  4. D4×10−7 m4 \times 10^{-7} \text{ m}

Correct answer

D. 4×10−7 m4 \times 10^{-7} \text{ m}

Step-by-step solution

The fringe width in Young's double slit experiment is given by β=λDd\beta = \dfrac{\lambda D}{d}

where λ\lambda is the wavelength, DD is the distance to the screen, and dd is the slit separation.

Given:
λ=100 pm=100×10−12 m=10−10 m\lambda = 100 \text{ pm} = 100 \times 10^{-12} \text{ m} = 10^{-10} \text{ m}
D=40 cm=0.4 mD = 40 \text{ cm} = 0.4 \text{ m}
β=0.1 mm=10−4 m\beta = 0.1 \text{ mm} = 10^{-4} \text{ m}

Rearranging the formula to solve for dd:
d=λDβd = \dfrac{\lambda D}{\beta}

Substituting the values:
d=10−10×0.410−4d = \dfrac{10^{-10} \times 0.4}{10^{-4}}

d=4×10−7 md = 4 \times 10^{-7} \text{ m}
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