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Ray Optics And Optical Instruments

Question

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Angular width of central maximum in the Fraunhoffer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength 6000 A^o. When the slit is illuminated by light of another wavelength, then the angular width decreases by 30%. The same decrease in angular width of central maximum is obtained when the original apparatus is immersed in a liquid. The refractive index of the liquid will be

1.25

1.42

1.67

1.5

Solution

B.

1.42

Given:-

λ₁ = 6000 A^o

λ₂ = ?

Angular width of central maximum = $\frac{2 λ}{a}$

θ₁ = $\frac{2 λ_{1}}{a}$ .....(i)

and θ₂ = $\frac{2 λ_{2}}{a}$

$∵$ θ₂ = θ₁ × 0.7

∴ θ₁ × 0.7 = $\frac{2 λ_{2}}{a}$ .....(ii)

From Eqs. (i) and (ii), we get

$\frac{1}{0.7} = \frac{λ_{1}}{λ_{2}}$

⇒ λ₂ = λ₂ × 0.7

= 4200 A^o

When immersed in liquid,

$λ_{2} = \frac{λ_{1}}{μ}$

$∵$ θ₂ = θ₁ × 0.7

⇒ $\frac{2 λ_{2}}{a} = \frac{2 λ_{1}}{a} \times 0.7$ $[λ_{2} = \frac{λ_{1}}{μ}]$

⇒ $\frac{2 \frac{λ_{1}}{μ}}{a} = \frac{2 λ_{1}}{a} \times 0.7$

⇒ $\frac{1}{μ}$ = 0.7

⇒ μ = 1.42

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