A thin prism of angle 15- made of glass of refractive index -1 - 1-5 is combined with another prism of glass of refractive index -2 - 1-75- The combination of the prism produces dispersion without deviation- The angle of the second prism should be-

Given: A_1=15^∘, nbsp; µ_1 = 1.5, nbsp; µ_2 = 1.75, δ=0 We know that δ =δ _1+δ _2 ⇒δ _1+δ _2=0 ⇒( μ _1 1)A_1+(μ _2 1)A_2=0 ⇒ A_2(1.75 1)= (1.5 1)15^∘ ⇒ A_2 ...

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Byju's Answer

Standard XII

Physics

Fringe Width in YDSE

A thin prism ...

Question

A thin prism of angle $15^{\circ}$ made of glass of refractive index $µ_{1} = 1.5$ is combined with another prism of glass of refractive index $µ_{2} = 1.75$ . The combination of the prism produces dispersion without deviation. The angle of the second prism should be:

12^{\circ}

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10^{\circ}

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7^{\circ}

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5^{\circ}

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Solution

The correct option is B $10^{\circ}$
Given: $A_{1} = 15^{\circ}$ , $µ_{1} = 1.5$ , $µ_{2} = 1.75$ , $δ = 0$

We know that
$δ = δ_{1} + δ_{2}$
$\Rightarrow δ_{1} + δ_{2} = 0$
$\Rightarrow (μ_{1} - 1) A_{1} + (μ_{2} - 1) A_{2} = 0$
$\Rightarrow A_{2} (1.75 - 1) = - (1.5 - 1) 15^{\circ}$
$\Rightarrow A_{2} = - \frac{0.5}{0.75} \times 15^{\circ}$

$∴ A_{2} = - 10^{\circ}$

Hence, $(B)$ is the correct answer.

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A thin prism of angle 15∘ made of glass of refractive index µ1=1.5 is combined with another prism of glass of refractive index µ2=1.75. The combination of the prism produces dispersion without deviation. The angle of the second prism should be:

The correct option is B 10∘Given: A1=15∘, µ1=1.5, µ2=1.75, δ=0 We know that δ=δ1+δ2 ⇒δ1+δ2=0 ⇒(μ1−1)A1+(μ2−1)A2=0 ⇒A2(1.75−1)=−(1.5−1)15∘ ⇒A2=−0.50.75×15∘ ∴A2=−10∘ Hence, (B) is the correct answer.

A thin prism of angle $15^{\circ}$ made of glass of refractive index $µ_{1} = 1.5$ is combined with another prism of glass of refractive index $µ_{2} = 1.75$ . The combination of the prism produces dispersion without deviation. The angle of the second prism should be: