to calculate the angle of minimum deviation (delta_m) for a prism.

Question

ScanSolve AI Tutor · Accepted Answer

The question asks to calculate the angle of minimum deviation ( $\delta_m$ ) for a prism.
Given:

Refractive index of the prism, $n = 1.4$
The "Reason" section specifies that it is an equilateral prism, which means the prism angle, $A = 60^\circ$ .

The formula for the refractive index of a prism in terms of the prism angle ( $A$ ) and the angle of minimum deviation ( $\delta_m$ ) is:

n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

Substitute the given values into the formula.
$1.4 = \frac{\sin\left(\frac{60^\circ + \delta_m}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)}$
Simplify the denominator.
$\sin\left(\frac{60^\circ}{2}\right) = \sin(30^\circ) = 0.5$
Substitute the simplified denominator back into the equation.
$1.4 = \frac{\sin\left(\frac{60^\circ + \delta_m}{2}\right)}{0.5}$
Isolate the sine term.
Multiply both sides by 0.5:
$1.4 \times 0.5 = \sin\left(\frac{60^\circ + \delta_m}{2}\right)$ $0.7 = \sin\left(\frac{60^\circ + \delta_m}{2}\right)$
Find the inverse sine (arcsin) of 0.7.
$\frac{60^\circ + \delta_m}{2} = \arcsin(0.7)$
Using a calculator, $\arcsin(0.7) \approx 44.427^\circ$ .
Solve for $\delta_m$ .
$\frac{60^\circ + \delta_m}{2} = 44.427^\circ$
Multiply both sides by 2:
$60^\circ + \delta_m = 2 \times 44.427^\circ$ $60^\circ + \delta_m = 88.854^\circ$
Subtract $60^\circ$ from both sides:
$\delta_m = 88.854^\circ - 60^\circ$ $\delta_m = 28.854^\circ$
Round the result to the nearest whole number.
$\delta_m \approx 29^\circ$
The calculated angle of minimum deviation is $29^\circ$ . This matches option A.

To calculate the angle of minimum deviation (delta_m) for a prism.