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NEET PHYSICSWave OpticsMedium

Question

The angular width of the central maximum in the Fraunhofer diffraction for λ=6000 A˚\lambda = 6000 \text{ \AA} is θ0\theta_0. When the same slit is illuminated by another monochromatic light, the angular width decreases by 30%30\%. The wavelength of this light is:

A

1800 A˚1800 \text{ \AA}

B

4200 A˚4200 \text{ \AA}

C

420 A˚420 \text{ \AA}

D

6000 A˚6000 \text{ \AA}

Step-by-Step Solution

The angular width of the central maximum in a single slit diffraction pattern is given by Δθ=2λa\Delta\theta = \frac{2\lambda}{a}, which implies that the angular width is directly proportional to the wavelength λ\lambda. Let the initial angular width be θ1=θ0\theta_1 = \theta_0 for λ1=6000 A˚\lambda_1 = 6000 \text{ \AA}. When the slit is illuminated by another monochromatic light, the new angular width θ2\theta_2 decreases by 30%30\%, so θ2=θ00.30θ0=0.70θ0\theta_2 = \theta_0 - 0.30\theta_0 = 0.70\theta_0. Using the direct proportionality (θλ\theta \propto \lambda), we have: θ2θ1=λ2λ1\frac{\theta_2}{\theta_1} = \frac{\lambda_2}{\lambda_1} 0.70θ0θ0=λ26000\frac{0.70\theta_0}{\theta_0} = \frac{\lambda_2}{6000} λ2=0.70×6000 A˚=4200 A˚\lambda_2 = 0.70 \times 6000 \text{ \AA} = 4200 \text{ \AA}. Thus, the wavelength of this light is 4200 A˚4200 \text{ \AA}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Wave Optics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWave Opticsangularcentralmaximumfraunhoferdiffraction

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