In given figure, S is a monochromatic point source emitting light of wavelength lambda = 500 nm. A t

Q: Why do the two lens halves act as coherent sources?

Because they both form images from the exact same original point source. Any phase changes in the original source are perfectly mirrored in both images, maintaining a constant phase difference between them.

Q: Why do we use the magnification formula to find the distance $d$?

The lens halves are shifted vertically. By finding the magnification, we can determine exactly how far the image is formed from the optic axis of that specific lens half. Adding the shift of the axis gives the absolute position.

Q: Why is the distance $D$ not $1.30\text{ m}$?

The distance $D$ in the interference formula is the distance from the *coherent sources* to the screen. Since the images $S_1$ and $S_2$ are formed $0.30\text{ m}$ to the right of the lens, the remaining distance to the screen is $1.30 - 0.30 = 1.0\text{ m}$.

Q: What does the third intensity maximum mean?

It refers to the bright fringe corresponding to a path difference of $3\lambda$. Its distance from the central maximum is exactly $3$ times the fringe width $\omega$.

Q: Why does reducing the gap increase the distance $OA$?

Reducing the gap between the lenses brings the two images $S_1$ and $S_2$ closer together, decreasing $d$. Since fringe width is inversely proportional to $d$, a smaller $d$ results in a wider fringe pattern, pushing the third maximum further out.

Question

In given figure, $S$ is a monochromatic point source emitting light of wavelength $\lambda = 500	ext{ nm}$. A thin lens of circular shape and focal length $0.10	ext{ m}$ 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 $0.5	ext{ mm}$. The distance along the axis from $S$ to $L_1$ and $L_2$ is $0.15	ext{ m}$ while that from $L_1$ and $L_2$ to $O$ is $1.30	ext{ m}$. The screen at $O$ is normal to $SO$.
(a) If the third intensity maximum occurs at the point $A$ on the screen, find the distance $OA$.
(b) If the gap between $L_1$ and $L_2$ is reduced from its original value of $0.5	ext{ mm}$, will the distance $OA$ increase, decrease, or remain the same.

Sigma Percentile · Accepted Answer

### The Setup: Billet's Split Lens

Imagine a point source of light placed in front of a lens that has been sliced perfectly in half, with the two halves slightly separated. This fascinating setup is a classic variation of Young's Double Slit Experiment, known as **Billet's split lens**.

Each half of the lens will refract the light and form its own independent image of the source. Because these two images are derived from the exact same original source, any phase changes in the original source are perfectly mirrored in both images. This means they will act as two perfectly coherent point sources, just like the two slits in Young's experiment.

### Finding the Coherent Sources

Let's find exactly where these images are formed. We use the standard lens formula:

$$ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} $$

The object distance $u$ is $-0.15	ext{ m}$, and the focal length $f$ is $+0.10	ext{ m}$. Substituting the values, we get:

$$ \frac{1}{v} - \frac{1}{-0.15} = \frac{1}{0.10} $$

$$ \frac{1}{v} = 10 - \frac{100}{15} = \frac{10}{3} $$

So, the image distance $v$ is exactly $0.30	ext{ m}$. Both images are formed at this distance to the right of the lens.

### The Geometry of the Interference

Next, we need the linear magnification to find the exact vertical positions of these images. Magnification $m$ is given by:

$$ m = \frac{v}{u} = \frac{0.30}{-0.15} = -2 $$

The negative sign tells us the images are inverted relative to the optic axis of each lens half. Look closely at the geometry. The optic axis of the top half is shifted up by $0.25	ext{ mm}$. Relative to this axis, the source is at $-0.25	ext{ mm}$. Multiplying by the magnification of $-2$, the image is formed $0.50	ext{ mm}$ above the axis. Adding the axis shift, $S_1$ is at $0.75	ext{ mm}$.

By symmetry, $S_2$ is at $-0.75	ext{ mm}$. The distance $d$ between them is:

$$ d = 0.75 - (-0.75) = 1.5	ext{ mm} $$

Now, what is the distance from our new coherent sources to the screen? The screen is at $1.30	ext{ m}$ from the lens, and the images are at $0.30	ext{ m}$. So, the effective distance $D$ is:

$$ D = 1.30	ext{ m} - 0.30	ext{ m} = 1.0	ext{ m} $$

### The Final Calculation

We have everything we need to find the fringe width $\omega$. Using the standard formula:

$$ \omega = \frac{\lambda D}{d} $$

We substitute the wavelength of $500	ext{ nm}$:

$$ \omega = \frac{(500 	imes 10^{-9})(1.0)}{1.5 	imes 10^{-3}} = \frac{1}{3}	ext{ mm} $$

The question asks for the position of the third intensity maximum. This occurs at a distance of three times the fringe width from the center:

$$ OA = 3\omega = 3 	imes \frac{1}{3}	ext{ mm} = 1	ext{ mm} $$

That's the answer to part (a)!

### What Happens When We Change the Gap?

For part (b), think about what happens if we reduce the gap between the lens halves. The distance $d$ between the images will decrease.

Since fringe width $\omega$ is inversely proportional to $d$, a smaller $d$ means the fringes will spread out. Therefore, the distance $OA$ to the third maximum will **increase**. A beautiful demonstration of wave optics!

Visualized Solution

\text{Billet's Split Lens Setup}

\text{Formation of Coherent Sources}

\text{Lens Formula Setup}

\text{Calculating Image Distance } v

\text{Linear Magnification } m

\text{Distance Between Sources } d

\text{Distance to Screen } D

\text{Fringe Width } \omega

\text{Position of Third Maximum}

\text{Effect of Reducing the Gap}

The Sigma Insight: Interference and Young's Double-Slit Experiment

The Setup

Finding the Coherent Sources

The Geometry of the Interference

The Final Calculation

What Happens When We Change the Gap?

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