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29 votes
29 votes
Light passes through a single slit of

width 8.77 x 10-6 m. The second (m = 2)
diffraction minimum occurs at an angle
of 5.62º. What is the wavelength of the
light IN NANOMETERS?

User Benjamin W
by
2.6k points

1 Answer

24 votes
24 votes

Final answer:

To find the wavelength of light in a single-slit diffraction where the second minimum occurs at an angle of 5.62° with a slit width of 8.77 x 10⁻⁶ m, use the formula λ = a sin(θ) / m, convert the angle to radians, solve for λ, and then convert from meters to nanometers.

Step-by-step explanation:

The student is asking about the wavelength of light in a single-slit diffraction experiment. The formula that relates the slit width λ, diffraction order m, angle of diffraction θ, and wavelength λ is given by: λ = a sin(θ) / m Where a is the slit width, and θ is the angle at which the m-th minimum occurs. Plugging in the values provided:a = 8.77 x 10-6 mm = 2θ = 5.62° First, we convert the angle θ to radians because the sine function requires the angle in radians:θ_radians = θ_degrees * (π / 180) Then, calculate the wavelength λ: λ = (8.77 x 10-6 m) * sin(5.62 * (π / 180)) / 2

After calculation, convert the wavelength from meters to nanometers by multiplying by 109.

User Rachelle Uy
by
3.0k points