Question

Red light (wavelength 632.8 nm in air) from a Helium-Neon laser is incident on a single slit of width 0.05 mm. The entire apparatus is immersed in water of refractive index 1.333. Determine the angular width of the central peak.

Answer 1

The angular width of the central peak in water when red light from a Helium-Neon laser is incident on a 0.05 mm wide slit is approximately 2.86°, after accounting for the change in wavelength due to the refractive index of water.

Step-by-step explanation:

The task is to determine the angular width of the central peak of a single-slit diffraction pattern in water when red light from a Helium-Neon laser is incident on the slit. The slit width is given as 0.05 mm and the refractive index of water as 1.333. The original wavelength of the Helium-Neon laser light in air is 632.8 nm. To find the angular width, the first step is to calculate the wavelength of the light in water by using the formula λ' = λ / n, where λ is the wavelength in air and n is the refractive index. This yields a new wavelength in water of 475.1 nm.

Next, using the formula for the angular width of the central peak in a single-slit diffraction pattern, θ = 2 × arctan(λ' / a), where λ' is the modified wavelength in water and a is the slit width, we calculate the angular width θ. Substituting the known values into the equation results in an angular width of approximately 2.86° for the central peak.