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The angular width of the central maximum in the Fraunhofer diffraction pattern of a slit is measured with light of wavelength 6000 angstrom. Case 1: When the slit is illuminated by light of a different wavelength, the angular width of the central maximum decreases by 30%. Case 2: When the original apparatus (6000 angstrom light) is immersed in a liquid, the angular width decreases by 30%. Which of the following statements are correct?

1. In Case 1, the wavelength of the new light is 4200 angstrom
2. In Case 2, the refractive index of the liquid is 10/7
3. In Case 1, the wavelength of the new light is 4200 angstrom AND in Case 2 the refractive index is 10/7
4. In Case 1 the new wavelength is 7800 angstrom AND in Case 2 the refractive index is 7/10

Correct answer: In Case 1, the wavelength of the new light is 4200 angstrom AND in Case 2 the refractive index is 10/7

Solution

Angular width W proportional to lambda/d. Case 1: W' = 0.70 W implies lambda'/d = 0.70 * 6000/d, so lambda' = 4200 angstrom. Case 2: Immersing in liquid changes effective wavelength to lambda/n. So (lambda/n)/d = 0.70 * lambda/d, giving 1/n = 0.70, so n = 10/7 approx 1.43. Both Case 1 and Case 2 results are correct in option C.

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