Prism Refractive Index and Minimum Deviation

The relationship between the refractive index ($n$) of a prism, its refracting angle ($A$), and the angle of minimum deviation ($\delta_{min}$) is given by the prism formula:

$n = \frac{\sin\left(\frac{A + \delta_{min}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$

Applying the Condition

The question states that the angle of minimum deviation is equal to the refracting angle:

$\delta_{min} = A$

Substitute this condition into the prism formula:

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

Simplify the expression:

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

Simplifying the Refractive Index Formula

Using the trigonometric identity $\sin(A) = 2 \sin\left(\frac{A}{2}\right) \cos\left(\frac{A}{2}\right)$, we get:

$n = \frac{2 \sin\left(\frac{A}{2}\right) \cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}$

This simplifies to:

$n = 2 \cos\left(\frac{A}{2}\right)$

Determining the Range of Refractive Index

For a prism, the refracting angle $A$ must satisfy $0^\circ < A < 180^\circ$. Consequently, the angle $\frac{A}{2}$ must satisfy $0^\circ < \frac{A}{2} < 90^\circ$.

In this range ($0^\circ < \frac{A}{2} < 90^\circ$), the cosine function, $\cos\left(\frac{A}{2}\right)$, varies between 0 and 1:

$0 < \cos\left(\frac{A}{2}\right) < 1$

Substituting this into the expression for $n$ ($n = 2 \cos\left(\frac{A}{2}\right)$):

$2 \times 0 < 2 \cos\left(\frac{A}{2}\right) < 2 \times 1$

$0 < n < 2$

However, the refractive index of any transparent material must be greater than 1 (i.e., $n > 1$). Combining this physical constraint with the derived inequality:

$1 < n < 2$

Therefore, the refractive index $n$ of the prism must be in the range $1 < n < 2$.