Physics Q40 – JEE-MAIN 2026

For a thin symmetric prism made of glass (refractive index 1.5), the ratio of incident angle and minimum deviation will be _______.

Clue 1 of 3: Thin Prism Approximation

For a thin prism, the angles of incidence, emergence, prism angle, and deviation are all considered small.

🧠 Full Solution Path

Step 1: Recall Formulas for a Thin Prism✦ Active

For a thin prism, the angle of minimum deviation $δ_{m}$ is given by:

δ_{m} = (n - 1) A

where $n$ is the refractive index and $A$ is the prism angle. For small angles, the incident angle $i$ at minimum deviation can be approximated using Snell's law ( $i = n r_{1}$ ). Since for minimum deviation $r_{1} = A / 2$ , we have:

i = n \frac{A}{2}

Step 2: Formulate the Ratio○ Expand

The required ratio is $\frac{i}{δ_{m}}$ . Substitute the expressions for $i$ and $δ_{m}$ from Step 1:

\frac{i}{δ_{m}} = \frac{n \frac{A}{2}}{(n - 1) A}

Simplify the expression by canceling $A$ :

\frac{i}{δ_{m}} = \frac{n}{2 (n - 1)}

Step 3: Calculate the Value○ Expand

Given the refractive index $n = 1.5$ , substitute this value into the ratio:

\frac{i}{δ_{m}} = \frac{1.5}{2 (1.5 - 1)} = \frac{1.5}{2 (0.5)} = \frac{1.5}{1} = \frac{3}{2}

Thus, the ratio of the incident angle to the minimum deviation is $3 : 2$ .

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 40 – JEE-MAIN 2026

Clue 1 of 3: Thin Prism Approximation

🧠 Full Solution Path