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Geometrical Optics question

2006 · Shift 0 · Q96

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Geometrical Optics
2006 · Shift 0 · Q96

Geometrical Optics question

2006 · Shift 0 · Q96

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1

The refractive index of a glass is

1.520

for red light and

1.525

for blue light. Let

{D_1}

and

{D_2}

be angles of minimum deviation for red and blue light respectively in a prism of this glass. Then,

A
${D_1} \lt {D_2}$
B
${D_1} = {D_2}$
C
${D_1}$ can be less than or greater than ${D_2}$ depending upon the angle of prism
D
${D_1} \gt {D_2}$

View written solutionFree

Correct answer: A

For a prism at minimum deviation, the refractive index is related to prism angle $A$ and minimum deviation $D_m$ by

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

So,

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

For a given prism, $A$ is fixed. Hence $\sin\left(\frac{A}{2}\right)$ is constant.

Therefore, if $n$ increases, then

\sin\left(\frac{A + D_m}{2}\right)

also increases.

Since the sine function is increasing for the physically relevant range of prism angles, it follows that

\frac{A + D_m}{2}

increases, and hence $D_m$ increases with $n$ .

Now given:

For red light: $n_r = 1.520$
For blue light: $n_b = 1.525$

Since

n_b > n_r,

we get

D_2 > D_1

because $D_2$ is the minimum deviation for blue light and $D_1$ for red light.

Thus,

D_1 < D_2

Checking options:

A: $D_1 < D_2$ Correct
B: $D_1 = D_2$ Incorrect
C: depends on prism angle Incorrect, for a fixed prism $D_m$ increases with $n$
D: $D_1 > D_2$ Incorrect

Therefore, the correct answer is A.

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