In an experiment used to demonstrate refraction, monochromatic light is transmitted through a transparent block with a refractive index of 1.4, which changes the speed of the light.If it's wavelength is 625 nm ( [math: 1nm=1×10−9m] 1nm = 1 imes 10^{-9} m ) in the block, determine the frequency of the light.Answers below are given in petaHertz, PHz ( [math: 1×1015Hz] 1 imes 10^{15} Hz ).

Question

In an experiment used to demonstrate refraction, monochromatic light is transmitted through a transparent block with a refractive index of 1.4, which changes the speed of the light.If it's wavelength is 625 nm ( [math: 1nm=1×10−9m] 1nm = 1 	imes 10^{-9} m  ) in the block, determine the frequency of the light.Answers below are given in petaHertz, PHz ( [math: 1×1015Hz] 1 	imes 10^{15} Hz ).

BlackTom AI · Accepted Answer

The problem asks for the frequency of monochromatic light after it passes into a transparent block with refractive index n = 1.4, given that the wavelength inside the block is 625 nm. The key principle is that the frequency of light remains unchanged when moving between media; only the speed and wavelength change.

Step 1: Relate speed, wavelength, and frequency in the block. The speed of light in the block is v = c/n, where c ≈ 3.00 × 10^8 m/s. With n = 1.4, v ≈ (3.00 × 10^8) / 1.4 ≈ 2.14 × 10^8 m/s.

Step 2: Use f = v / λ in the block. The wavelength in the block is given as λ_block = 625 nm = 625 × 10^-9 m. Therefore, f = v / λ_block ≈ (2.14 × 10^8) / (625 × 10^-9) Hz.

Step 3: Compute the frequency. (2.14 × 10^8) / (625 × 10^-9) = (2.14 × 10^8) × (1.6 × 10^6) ≈ 3.43 × 10^14 Hz.

Step 4: Express the result in Petahertz. 1 PHz = 10^15 Hz, so f ≈ 0.343 PHz.

Note on interpretation: Another way to see this is that the frequency is invariant across media, so you can also write f = c / λ0, where λ0 is the wavelength in vacuum. Since λ_block = λ0 / n, we have λ0 = n × λ_block = 1.4 × 625 nm = 875 nm, and f = c / λ0 ≈ 3.00 × 10^8 / 875 × 10^-9 ≈ 3.43 × 10^14 Hz, which again is about 0.343 PHz.

Potential source of confusion: If one incorrectly uses the given 625 nm as the vacuum wavelength, one would obtain f = c / λ0 ≈ 0.48 PHz. However, the problem states the wavelength inside the block is 625 nm, so the correct calculation must account for the refractive index when relating to the vacuum wavelength. The consistent result with the given λ_block is approximately 0.343 PHz.

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