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Cxamples 1. If in at prism, \( \mathrm{A}=60^{\circ} \) and if \( \mathrm{m}=1.52 \), calculate the minimum deviation of a ray of light that posses through it.

Ask by Kirk Hall. in Ghana
Mar 15,2025

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Answer

The minimum deviation of the ray is approximately \( 38.92^\circ \).

Solution

1. In a prism, the minimum deviation \( \delta_{\text{min}} \) is given by the formula \[ \delta_{\text{min}} = 2\arcsin\left(n \sin\frac{A}{2}\right) - A \] where - \( A \) is the prism’s apex angle, - \( n \) is the refractive index of the prism material. 2. Given: \[ A = 60^\circ,\quad n = 1.52. \] 3. Calculate \( \sin\frac{A}{2} \): Since \( \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ \) and \[ \sin 30^\circ = 0.5, \] we have \[ n \sin \frac{A}{2} = 1.52 \times 0.5 = 0.76. \] 4. Compute the value of \( \arcsin(0.76) \) (in degrees): \[ \arcsin(0.76) \approx 49.46^\circ. \] 5. Substitute into the formula for \( \delta_{\text{min}} \): \[ \delta_{\text{min}} = 2 \times 49.46^\circ - 60^\circ \approx 98.92^\circ - 60^\circ = 38.92^\circ. \] 6. Therefore, the minimum deviation of the ray is approximately \[ \delta_{\text{min}} \approx 38.92^\circ. \]

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To find the minimum deviation, we use the formula for a prism: \( D = (n - 1)A \), where \( D \) is the minimum deviation, \( n \) is the refractive index, and \( A \) is the angle of the prism. Plugging in the values, we have \( D = (1.52 - 1) \times 60^{\circ} = 0.52 \times 60^{\circ} = 31.2^{\circ} \). Therefore, the minimum deviation of the ray of light as it passes through the prism is \( 31.2^{\circ} \). In practical terms, understanding the minimum deviation can help in designing optical devices like prisms used in binoculars and cameras. By knowing the refractive index and the prism angle, engineers can optimize the paths of light to achieve clearer images and better vision. So, the next time you snap a picture or look through binoculars, remember that those beautiful views depend on the magic of minimum deviation!

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