StemCET Logo

Physics Question 40 – JEE-MAIN 2026

← Back
A ray of light passing through an equilateral prism is having velocity 2.12×108 m/s in the prism material, then the minimum angle of deviation is _______ degrees.

Understand how the speed of light in a medium relates to its refractive index, and recall that an equilateral prism has a prism angle of 60∘.

Step 1: Calculate the Refractive Index✦ Active

The refractive index n of the prism material is given by the ratio of the speed of light in vacuum (c=3×108 m/s) to the speed of light in the prism material (v=2.12×108 m/s).

n=cv=3×108 m/s2.12×108 m/s≈1.415
Step 2: Apply the Minimum Deviation Formula○ Expand

For a prism, the refractive index n is related to the prism angle A and the minimum angle of deviation δm by the formula:

n=sin⁡(A+δm2)sin⁡(A2)

For an equilateral prism, the prism angle A=60∘. Substituting the values:

1.415=sin⁡(60∘+δm2)sin⁡(60∘2)=sin⁡(60∘+δm2)sin⁡(30∘)
💡 Teacher's Secret Hint

Recall that sin⁡(30∘)=0.5.

Step 3: Solve for Minimum Deviation○ Expand

Substitute sin⁡(30∘)=0.5 into the equation from Step 2:

1.415=sin⁡(60∘+δm2)0.5

Rearranging to solve for the sine term:

sin⁡(60∘+δm2)=1.415×0.5=0.7075

We know that sin⁡(45∘)≈0.7071. Therefore:

60∘+δm2≈45∘

Solving for δm:

60∘+δm=90∘
δm=90∘−60∘=30∘
💡 Teacher's Secret Hint

Remember the common trigonometric values for angles like 30∘ and 45∘.

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