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.

Question

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.

Answer 1

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:b = λ / (2 * sin(α))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.Given that the wavelength λ is 600 nm and the distance b is 4 mm, we can rearrange the formula to solve for α:sin(α) = λ / (2 * b)α = arcsin(λ / (2 * b))Substituting the given values:α = arcsin(600 nm / (2 * 4 mm))α ≈ 0.075 radiansTherefore, the angle α between the surfaces of the wedge II is approximately 0.075 radians.

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.

Решения

Ответ