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B3. Monochromatic light (lambda =600nm) is incident on a th to its surface. Determine the angle alpha between the surfaces of the wedge II distance b between neighboring interference maxima in the reflected light is 4MM.

Вопрос

B3. Monochromatic light (lambda =600nm) is incident on a th
to its surface. Determine the angle alpha  between the surfaces of the wedge II
distance b between neighboring interference maxima in the reflected light is
4MM.

B3. Monochromatic light (lambda =600nm) is incident on a th to its surface. Determine the angle alpha between the surfaces of the wedge II distance b between neighboring interference maxima in the reflected light is 4MM.

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To determine the angle α between the surfaces of the wedge II, we can use the formula for the distance between neighboring interference maxima in the reflected light:<br /><br />b = λ / (2 * sin(α))<br /><br />where b is the distance between neighboring interference maxima, λ is the wavelength of the incident light, and α is the angle between the surfaces of the wedge.<br /><br />Given that the wavelength λ is 600 nm and the distance b is 4 mm, we can rearrange the formula to solve for α:<br /><br />sin(α) = λ / (2 * b)<br /><br />α = arcsin(λ / (2 * b))<br /><br />Substituting the given values:<br /><br />α = arcsin(600 nm / (2 * 4 mm))<br /><br />α ≈ 0.075 radians<br /><br />Therefore, the angle α between the surfaces of the wedge II is approximately 0.075 radians.
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