Question

In the figure shown $S$ is a monochromatic point source emitting light of wavelength $= 500 nm$ . A thin lens of circular shape and focal length $0 .10 m$ is cut into two identical halves $L_{1}$ and $L_{2}$ by a plane passing through a diameter. The two halves are placed symmetrically about the central axis SO with a gap of $0.5 mm$ . The distance along the axis from $S$ to $L_{1}$ and $L_{2}$ is $0.15 m$ , while that from $L_{1}$ and $L_{2}$ to $O$ is $1.30 m$ .

The screen at $O$ is normal to SO. (i) If the third intensity maximum occurs at the point $A$ on the screen, find the distance OA. (ii) If the gap between $L_{1}$ and $L_{2}$ is reduced from its original value of $0.5 mm$ , will the distance OA increase decrease or remain the same?

Accepted Answer

(i) For the lens, $u = - 0.15 m; f = + 0.10 m$

Therefore, using $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ we have

$\frac{1}{v} = \frac{1}{u} + \frac{1}{f} = \frac{1}{(- 0.15)} + \frac{1}{(0.10)}$ or $v = 0.3 m$

Linear magnification, $m = \frac{v}{u} = \frac{0.3}{- 0.15} = - 2$

Hence, two images $S_{1}$ and $S_{2}$ of $S$ will be formed at $0.3 m$ from the lens as shown in figure. Image $S_{1}$ due to part $1$ will be formed at $0.5 m m$ above its optic axis $(m = - 2)$ . Similarly, $S_{2}$ due to part $2$ is formed $0.5 mm$ below the optic axis of this part as shown.

Hence, $d =$ distance between $S_{1}$ and $S_{2} = 1.5 mm$

$Δ = 1.30 - 0.30 = 1.0 m = 10^{3} mm$

$0.5 mm$

Therefore, fringe width,

$ω = \frac{λ D}{d} = \frac{(5 \times 10^{- 4}) (10^{3})}{(1.5)} mm = \frac{1}{3} mm$

Now, as the point $A$ is at the third maxima

$O A = 3 ω = 3 (\frac{1}{3}) mm or O A = 1 mm$

(ii) If the gap between $L_{1}$ and $L_{2}$ is reduced, $d$ will decrease. Hence, the fringe width $ω$ will increase or the distance $O A$ will increase.

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