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Chapter 4: Problem 103

Calculate the wavelength of light that produces its first minimum at an angle of \(36.9^{\circ}\) when falling on a single slit of width \(1.00 \mu \mathrm{m}\)

Short Answer

Expert verified
The wavelength of light that produces its first minimum at an angle of \(36.9^{\circ}\) when falling on a single slit of width \(1.00\mu \mathrm{m}\) is approximately 438 nm.

Step by step solution

01

Identify the given information

In this exercise, we are given: - Angle of the first minimum: \(36.9^{\circ}\) - Width of the slit: \(1.00 \mu \mathrm{m}\) We are asked to find the wavelength of light.
02

Convert the angle to radians

All trigonometric functions in this problem should be calculated in radians. Therefore, we need to convert the given angle from degrees to radians. angle (in radians) = angle (in degrees) × \(\frac{\pi}{180}\) angle (in radians) = \(36.9^{\circ} \times \frac{\pi}{180}\) angle (in radians) = 0.644 radians
03

Apply the formula for single-slit diffraction

The formula for single-slit diffraction is given by: sin(θ) = \(\frac{m\lambda}{a}\) where - \(a\) is the width of the slit, - \(\lambda\) is the wavelength of the light, - \(m\) is the order of the minimum (in this case, \(m=1\) for the first minimum), and - θ is the angle of the first minimum in radians. We are solving for the wavelength \(\lambda\), so the formula can be rearranged as: \(\lambda = \frac{a \cdot \sin(\theta)}{m}\)
04

Substitute the given values into the formula

Now we can plug in the given values and solve for the wavelength \(\lambda\). \(\lambda = \frac{1.00 \times 10^{-6} m \cdot \sin(0.644)}{1}\) \(\lambda \approx 4.38 \times 10^{-7} m\)
05

Express the result in nanometers

To express the result in nanometers, we need to multiply the wavelength in meters by \(10^9\). \(\lambda \approx 4.38 \times 10^{-7} m \times 10^9 \frac{nm}{m}\) \(\lambda \approx 438 nm\) So, the wavelength of light that produces its first minimum at an angle of \(36.9^{\circ}\) when falling on a single slit of width \(1.00\mu \mathrm{m}\) is approximately 438 nm.

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