Can we use the prism formula $\mu = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)}$ directly?

Yes! That formula is derived from Snell's law and the symmetry condition. You can plug in $\mu = 1.5$ and $A = 60^\circ$ to find $\delta_m$, and then use $i = (A+\delta_m)/2$, which gives the exact same result.

JEE Advanced 1998

LEVELJEE Advanced

Animated Solution for Physics - Optics: A prism of refractive index $n_{1}$ and another prism of refractive index $n_{2}$ are stuck together with a gap as shown in the figure. The angles of the prism are as shown. $n_{1}$ and $n_{2}$ depend on $λ$ , the wavelength of light according to : $n_{1} = 1.20 + \frac{10.8 \times 1 0 ^{4}}{λ ^{2}}$ and $n_{2} = 1.45 + \frac{1.80 \times 1 0 ^{4}}{λ ^{2}}$ where $λ$ is in nm. (a) Calculate the wavelength $λ_{0}$ for which rays incident at any angle on the interface $B C$ pass through without bending at that interface. (b) For light of wavelength $λ_{0}$ , find the angle of incidence $i$ on the face $A C$ such that the deviation produced by the combination of prisms is minimum.

Q: Why does the ray not bend when $n_1 = n_2$?

According to Snell's law, $n_1 \sin i = n_2 \sin r$. If $n_1 = n_2$, then $\sin i = \sin r$, which means the angle of incidence equals the angle of refraction. The ray travels in a straight line without any deviation.

Visualized Solution

$n_{1} = n_{2}$

For the ray to pass through the interface $B C$ without any deviation for all angles of incidence, the refractive indices of the two prisms must be identical.
$n_{1} = n_{2}$

$1.20 + \frac{10.8 \times 1 0 ^{4}}{λ _{0}^{2}} = 1.45 + \frac{1.80 \times 1 0 ^{4}}{λ _{0}^{2}}$

Substitute the given expressions for $n_{1}$ and $n_{2}$ :
$1.20 + \frac{10.8 \times 1 0 ^{4}}{λ _{0}^{2}} = 1.45 + \frac{1.80 \times 1 0 ^{4}}{λ _{0}^{2}}$

$λ_{0} = 600 nm$

Rearranging the terms to solve for $λ_{0}$ :
$\frac{10.8 \times 1 0 ^{4} - 1.80 \times 1 0 ^{4}}{λ _{0}^{2}} = 1.45 - 1.20$
$\frac{9.0 \times 1 0 ^{4}}{λ _{0}^{2}} = 0.25$
$λ_{0}^{2} = \frac{9.0 \times 1 0 ^{4}}{0.25} = 36 \times 1 0^{4}$
$λ_{0} = 600 nm$

$n = 1.50$

Calculate the common refractive index $n$ at $λ_{0} = 600 nm$ :
$n = 1.20 + \frac{10.8 \times 1 0 ^{4}}{( 600 ) ^{2}}$
$n = 1.20 + \frac{108000}{360000} = 1.20 + 0.30 = 1.50$

$A = 6 0^{\circ}$

Since $n_{1} = n_{2} = 1.5$ , the interface $B C$ optically disappears. The combination acts as a single equilateral prism with an effective prism angle $A = 6 0^{\circ}$ .

$r_{1} = 3 0^{\circ}$

For minimum deviation, the ray passes symmetrically through the prism. The angle of refraction $r_{1}$ at the first face is:
$r_{1} = \frac{A}{2} = \frac{6 0 ^{\circ}}{2} = 3 0^{\circ}$

$i = sin^{- 1} (\frac{3}{4})$

Apply Snell's law at the first face $A C$ :
$sin i = n sin r_{1}$
$sin i = 1.5 \times sin 3 0^{\circ} = 1.5 \times 0.5 = 0.75 = \frac{3}{4}$
$i = sin^{- 1} (\frac{3}{4})$

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The Sigma Insight: Refraction and Dispersion through Prism

Solution Diagram

Imagine you have two prisms, glued together but with a tiny gap between them. The first part of the question asks for the specific wavelength of light where the interface between these two prisms effectively 'disappears'.

The Disappearing Interface For a light ray to pass through the interface $B C$ without any bending, regardless of its angle of incidence, the optical density of both media must be identical

This means their refractive indices must be equal.

We set the given equations for

n_{1}

and

n_{2}

equal to each other:

1.20 + \frac{10.8 \times 1 0 ^{4}}{λ _{0}^{2}} = 1.45 + \frac{1.80 \times 1 0 ^{4}}{λ _{0}^{2}}

By rearranging the terms to solve for

λ_{0}

, we get:

\frac{9.0 \times 1 0 ^{4}}{λ _{0}^{2}} = 0.25

Solving this gives

λ_{0}^{2} = 36 \times 1 0^{4}

, which means the wavelength

λ_{0}

is exactly $600 nm$ .

The Combined Prism Now, let's find the actual refractive index at this wavelength

Plugging

λ_{0} = 600 nm

back into the equation for

n_{1}

n = 1.20 + \frac{10.8 \times 1 0 ^{4}}{( 600 ) ^{2}} = 1.20 + 0.30 = 1.50

Since both prisms now share the exact same refractive index of

1.50

, the boundary between them vanishes optically. The entire setup behaves as a single, uniform equilateral prism. For this combined system, the effective prism angle

A

6 0^{\circ}

Minimum Deviation For the condition of minimum deviation, the light ray must travel symmetrically through the prism

This symmetry dictates that the angle of refraction at the first face,

r_{1}

, is exactly half of the prism angle:

r_{1} = \frac{A}{2} = \frac{6 0 ^{\circ}}{2} = 3 0^{\circ}

Finally, we apply Snell's law at the entry face

A C

to find the angle of incidence

i

sin i = n sin r_{1}

sin i = 1.5 \times sin 3 0^{\circ} = 1.5 \times 0.5 = 0.75 = \frac{3}{4}

Therefore, the required angle of incidence is $i = sin^{- 1} (\frac{3}{4})$ .

Common Traps & FAQs

Why does the ray not bend when

n_{1} = n_{2}

According to Snell's law,

n_{1} sin i = n_{2} sin r

. If

n_{1} = n_{2}

, then

sin i = sin r

, which means the angle of incidence equals the angle of refraction. The ray travels in a straight line without any deviation.

Why do we assume the combined prism angle is

6 0^{\circ}

The problem implies that the outer faces of the combined system form an equilateral prism. When the interface optically disappears, the light only interacts with the outer faces, which are oriented at

6 0^{\circ}

to each other.

What happens to the gap between the prisms?

The gap is considered to be an infinitesimally thin parallel-sided air film. A parallel-sided film does not change the angular deviation of a ray passing through it; it only causes a negligible lateral shift.

Why is

r_{1} = A /2

for minimum deviation?

At minimum deviation, the light ray passes symmetrically through the prism. This symmetry means the angle of refraction at the first face equals the angle of incidence at the second face, leading to

r_{1} = r_{2} = A /2

Can we use the prism formula

μ = \frac{s i n (( A + δ _{m} ) /2 )}{s i n ( A /2 )}

directly?

Yes! That formula is derived from Snell's law and the symmetry condition. You can plug in

μ = 1.5

and

A = 6 0^{\circ}

to find

δ_{m}

, and then use

i = (A + δ_{m}) /2

, which gives the exact same result.

Visualized Solution

$n_{1} = n_{2}$

$1.20 + \frac{10.8 \times 1 0 ^{4}}{λ _{0}^{2}} = 1.45 + \frac{1.80 \times 1 0 ^{4}}{λ _{0}^{2}}$

$λ_{0} = 600 nm$

$n = 1.50$

$A = 6 0^{\circ}$

$r_{1} = 3 0^{\circ}$

$i = sin^{- 1} (\frac{3}{4})$

The Sigma Insight: Refraction and Dispersion through Prism

The Disappearing Interface For a light ray to pass through the interface $B C$ without any bending, regardless of its angle of incidence, the optical density of both media must be identical

The Combined Prism Now, let's find the actual refractive index at this wavelength

Minimum Deviation For the condition of minimum deviation, the light ray must travel symmetrically through the prism

Similar Questions

A thin prism $P_{1}$ with angle $4^{\circ}$ and made from glass of refractive index $1.54$ is combined with another thin prism $P_{2}$ made from glass of refractive index $1.72$ to produce dispersion without deviation. The angle of the prism $P_{2}$ is

An isosceles prism of angle $12 0^{\circ}$ has a refractive index $1.44$ . Two parallel rays of monochromatic light enter the prism parallel to each other in air as shown. The rays emerging from the opposite face

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n1​=n2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1.20+λ02​10.8×104​=1.45+λ02​1.80×104​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ0​=600 nm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=1.50

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A=60∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r1​=30∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i=sin−1(43​)

The Sigma Insight: Refraction and Dispersion through Prism

The Combined Prism Now, let's find the actual refractive index at this wavelength

Minimum Deviation For the condition of minimum deviation, the light ray must travel symmetrically through the prism

Similar Questions

$n_{1} = n_{2}$

$1.20 + \frac{10.8 \times 1 0 ^{4}}{λ _{0}^{2}} = 1.45 + \frac{1.80 \times 1 0 ^{4}}{λ _{0}^{2}}$

$λ_{0} = 600 nm$

$n = 1.50$

$A = 6 0^{\circ}$

$r_{1} = 3 0^{\circ}$

$i = sin^{- 1} (\frac{3}{4})$