SATHEE: Optics 7 Question 43

Optics 7 Question 43

44. Two parallel beams of light $P$ and $Q$ (separation $d$ ) containing radiations of wavelengths $4000 \AA$ and $5000 \AA$ (which are mutually coherent in each wavelength separately) are incident normally on a prism as shown in figure.

The refractive index of the prism as a function of wavelength is given by the relation, $μ (λ) = 1.20 + \frac{b}{λ^{2}}$ where $λ$ is in $\AA$ and $b$ is positive constant. The value of $b$ is such that the condition for total reflection at the face $A C$ is just satisfied for one wavelength and is not satisfied for the other.

$(1991, 2 + 2 + 4 M)$

(a) Find the value of $b$ .

(b) Find the deviation of the beams transmitted through the face $A C$ .

(c) A convergent lens is used to bring these transmitted beams into focus. If the intensities of the upper and the lower beams immediately after transmission from the face $A C$ , are $4 I$ and $I$ respectively, find the resultant intensity at the focus.

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Answer:

Correct Answer: 44. (a) $b = 8 \times 10^{5} (\AA)^{2}$ (b) $δ_{4000 \AA} = 37^{\circ}, δ_{5000 \AA} = {27.13}^{\circ}$ (c) $9 I$

Solution:

(a) Total internal reflection (TIR) will take place first for that wavelength for which critical angle is small or $μ$ is large.

From the given expression of $μ$ , it is more for the wavelength for which value

of $λ$ is less.

Thus, condition of TIR is just satisfied for $4000 \AA$ .

or $i = θ_{c}$ for $4000 \AA$ or $θ = θ_{c}$ or $\sin θ = \sin θ_{c}$

$\begin{aligned} or 0.8 = \frac{1}{μ} \\ or 0.8 = \frac{1}{1.20 + \frac{b}{(4000)^{2}}} \end{aligned}$

Solving this equation, we get $b = 8.0 \times 10^{5} (\AA)^{2}$

(b)

For, $4000 \AA$ condition of TIR is just satisfied. Hence, it will emerge from $A C$ , just grazingly.

or $δ_{4000 \AA} = 90^{\circ} - i = 90^{\circ} - \sin^{- 1} (0.8) \approx 37^{\circ}$

For $5000 \AA μ = 1.2 + \frac{b}{λ^{2}} = 1.2 + \frac{8.0 \times 10^{5}}{(5000)^{2}} = 1.232$

Applying $μ = \frac{\sin i_{air}}{\sin i_{medium}}$ or $1.232 = \frac{\sin i_{air}}{\sin θ}$

$\begin{aligned} = \frac{\sin i_{air}}{0.8} \Rightarrow i_{air} = {80.26}^{\circ} \\ ∴ δ_{5000 \AA} & = i_{air} - i_{medium} \\ = {80.26}^{\circ} - \sin^{- 1} (0.8) = {27.13}^{\circ} \end{aligned}$

(c)

Path difference between rays 1 and 2

$Δ x = μ (Q S) - P R$

Further, $\frac{Q S}{P S} = \sin i \Rightarrow \frac{P R}{P S} = \sin r$

$∴ \frac{P R / P S}{Q S / P S} = \frac{\sin r}{\sin i} = μ \Rightarrow μ (Q S) = P R$

Substituting in Eq. (i), we get $Δ x = 0$ .

$∴$ Phase difference between rays 1 and 2 will be zero.

Or these two rays will interfere constructively. So, maximum intensity will be obtained from their interference.

$or$

$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = (\sqrt{4 I} + \sqrt{I})^{2} = 9 I$

NOTE In this question we have written,

$μ = \frac{\sin r}{\sin i} not \frac{\sin i}{\sin r}$

because in medium angle with normal is $i$ and in air angle with normal is $r$ .

$μ = \frac{\sin i_{air}}{\sin i_{medium}}$

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Optics 7 Question 43

44. Two parallel beams of light $P$ and $Q$ (separation $d$ ) containing radiations of wavelengths $4000 \AA$ and $5000 \AA$ (which are mutually coherent in each wavelength separately) are incident normally on a prism as shown in figure.

Answer:

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Optics 7 Question 43

44. Two parallel beams of light P and Q (separation d ) containing radiations of wavelengths 4000\AA and 5000\AA (which are mutually coherent in each wavelength separately) are incident normally on a prism as shown in figure.

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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44. Two parallel beams of light $P$ and $Q$ (separation $d$ ) containing radiations of wavelengths $4000 \AA$ and $5000 \AA$ (which are mutually coherent in each wavelength separately) are incident normally on a prism as shown in figure.