Question

A lens of diameter

5.0 cm

and focal length

f = 25.0 cm

was cut along the diameter into two identical halves. In the process, the layer of the lens,

a = 1.00 mm

in thickness, was lost. Then the halves were put together to form a composite lens. In this focal plane, a narrow slit was placed, emitting monochromatic light with wavelength

λ = 0.60 μm

. Behind the lens a screen was located at a distance

b = 50 cm

from it. Find:
(a) the width of a fringe on the screen and the number of possible maxima;
(b) the maximum width of the slit

δ_{\max}

at which the fringes on the screen will be still observed sufficiently sharp.

Accepted Answer

When the two halves were put together to form a composite lens, then the optical axis of the upper half is shifted downward by a distance $\frac{a}{2} .$ In the similar fashion, the optical axis of the lower part is shifted upward by a distance $f = 25.0 cm$

The angle of deviation produced by the upper or the lower part of the lens, in incident ray is

$ψ = \frac{a}{2 f}$

So, the total angle of deviation produced is,

$2 ψ = \frac{a}{f}$

The fringe width is given by,

$β = \frac{λ}{2 ψ} = \frac{f λ}{a} . . . (1)$

Here, $f = 25 cm, λ = 0.6 \times 10^{- 6} m, a = 1 mm$

Putting the values, we get the fringe width, $β = 0.15 mm$

Since the arrangement is similar to Young's experiment, so the number of fringes $= 2 [n_{1}] + 1$

where $n_{1} =$ Number of fringes on the either side (between $A$ and $B$ ) of the central point $C$ of the screen.

Let $A C = B C = l$

$∴ l = b \tan ψ = b ψ = \frac{b a}{2 f}$

$∴ n_{1} = \frac{l}{β} = \frac{b a}{2 f \times \frac{f λ}{a}}$

$∴ n_{1} = \frac{a^{2} b}{2 f^{2} λ} . . . (2)$

Putting the values, we get,

$n_{1} = \frac{{(10^{- 3})}^{2} \times (50 \times 10^{- 2})}{2 \times (25 \times 10^{- 2}) \times 0.6 \times 10^{- 6}} = 6.67$

$∴ n = 2 [n_{1}] + 1 = 2 \times 6 + 1 = 13$

(b) The fringe pattern observed on the screen will not be sharp if the pattern due to the two narrow slits are $\frac{β}{2}$ (half of the fringe width) apart, because then the maxima due to one of the slits will overlap with the minima of the other slit.

$∵ \tan ∆ θ = \frac{∆ y}{b} = \frac{δ}{f}$

$δ = \frac{∆ y}{b} f$

$∴ ∆ y = \frac{b δ}{f}$

For maximum width of the slit, we have,

$∵ \frac{b δ_{\max}}{f} = \frac{β}{2}$

$\frac{b δ_{\max}}{f} = \frac{f λ}{2 a}$

$δ_{\max} = \frac{f^{2} λ}{2 b a}$

$∴ δ_{\max} = \frac{f^{2} λ}{2 b a}$

On putting the values, $δ_{\max} = 37 μm$ .

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