A double-slit interference experiment is performed in the air. Later, the same apparatus is immersed in benzene (which has an index of refraction of 1.50), and the experiment is repeated. When the apparatus is in benzene, you observe that the interference fringes are

Question

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When a double-slit interference setup is moved from air into a medium with a higher refractive index, the wavelength of light inside the medium becomes shorter because λ = λ0 / n, where λ0 is the wavelength in vacuum/air and n is the refractive index of the medium. The fringe spacing on the screen is Δy ≈ (λ L)/d, so as λ decreases (due to n increasing from ~1 to 1.50), the fringe spacing Δy also decreases.

Option 1: "more closely spaced than when the apparatus is in air." This follows from the reduction of wavelength in benzene; since Δy scales with λ, a smaller λ yields a smaller Δy, meaning fringes are closer together.

Option 2: "equally spaced as when the apparatus is in air." This would be true only if the wavelength and the optical path difference were unchanged by the medium. However, the medium changes λ by a factor of 1/n, so the spacing is not equal.

Option 3: "more widely spaced than when the apparatus is in air." A larger spacing would require a larger wavelength inside the medium. But in a medium with n > 1, the wavelength is reduced, not increased, so the fringes cannot be more widely spaced.

In summary, the key concept is that the fringe separation depends on the wavelength inside the medium, which becomes shorter as the refractive index rises, leading to tighter (more closely spaced) fringes on the screen.

A double-slit interference experiment is performed in the air. Later, the same apparatus is immersed in benzene (which has an index of refraction of 1.50), and the experiment is repeated. When the apparatus is in benzene, you observe that the interference fringes are单项选择题

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