A slit of width d is placed in front of a lens of focal length 0.5 m and is illuminated normally wit

Q: Why do we use a lens in this setup?

The lens is used to focus the parallel diffracted rays onto a screen at a finite distance. Without the lens, the screen would theoretically need to be placed at infinity to observe a sharp Fraunhofer diffraction pattern.

Q: Why is the condition for a minimum $d \sin \theta = \lambda$ and not a maximum like in double slit?

In a single slit, we divide the slit into two halves. For the first minimum, the path difference between a ray from the top edge and a ray from the middle is $\lambda/2$, causing destructive interference. This pairs up all rays across the slit, resulting in a net zero intensity when $d \sin \theta = \lambda$.

Q: Can we always use the small angle approximation?

Yes, in typical optical diffraction setups, the wavelength of light is much smaller than the slit width ($d \gg \lambda$), making the diffraction angle $\theta$ very small. For small angles, $\sin \theta \approx \tan \theta \approx \theta$ in radians.

Q: What happens if we use white light instead of monochromatic light?

The central maximum will be white because all wavelengths have zero path difference there. However, the minima and secondary maxima will be colored and spread out, with red light diffracting at a wider angle than blue light.

Q: Why is the distance to the screen taken as the focal length $f$?

Because the slit is placed in front of the lens and illuminated normally, the diffracted rays at any specific angle $\theta$ are parallel to each other. A converging lens focuses parallel rays exactly at its focal plane.

Question

A slit of width $d$ is placed in front of a lens of focal length $0.5	ext{ m}$ and is illuminated normally with light of wavelength $5.89 	imes 10^{-7}	ext{ m}$. The first diffraction minima on either side of the central diffraction maximum are separated by $2 	imes 10^{-3}	ext{ m}$. The width $d$ of the slit is ...... $	ext{m}$.

Sigma Percentile · Accepted Answer

Have you ever wondered what happens when light passes through a tiny opening? Instead of a sharp shadow, it spreads out, creating a beautiful pattern of bright and dark bands. This phenomenon is known as **diffraction**. In this problem, we are going to act like optical detectives to find the exact width of a slit, just by looking at the pattern it creates!

### Analyzing the Setup
Imagine a narrow slit of width $d$. We shine a monochromatic light of wavelength $\lambda = 5.89 	imes 10^{-7}	ext{ m}$ directly onto it. Right behind the slit, we place a converging lens with a focal length $f = 0.5	ext{ m}$.

Why the lens? When light diffracts through the slit, the rays spread out at various angles. The lens takes all the parallel rays diffracting at a specific angle $	heta$ and focuses them onto a single point on a screen placed at its focal plane. This means the distance from the lens to the screen is exactly $f$.

### The Master Equation
The diffraction pattern on the screen consists of a broad, bright central maximum, flanked by alternating dark and bright fringes. The dark fringes, or **minima**, occur where the light waves from different parts of the slit perfectly cancel each other out (destructive interference).

The condition for the *first* minimum is elegantly simple:
$$d \sin	heta = \lambda$$

Here, $	heta$ is the angle at which the first minimum occurs.

### The Small Angle Approximation
In most practical setups, the wavelength of light is much smaller than the slit width, making the angle $	heta$ very small. For small angles, we can use a handy mathematical trick: $\sin	heta \approx 	an	heta$.

If we look at the geometry of our setup, the tangent of the angle $	heta$ is simply the opposite side over the adjacent side. The opposite side is the distance $y$ from the central maximum to the first minimum, and the adjacent side is the focal length $f$.
$$\sin	heta \approx 	an	heta = \frac{y}{f}$$

Substituting this back into our master equation, we get:
$$d \left(\frac{y}{f}ight) = \lambda$$
$$d = \frac{\lambda f}{y}$$

### Final Calculation
The problem states that the distance between the first minima on *either side* of the central maximum is $2 	imes 10^{-3}	ext{ m}$. This total distance is $2y$.
$$2y = 2 	imes 10^{-3}	ext{ m}$$
$$y = 1 	imes 10^{-3}	ext{ m}$$

Now, we have all the pieces of the puzzle! Let's plug them into our rearranged equation:
$$d = \frac{(5.89 	imes 10^{-7}	ext{ m}) 	imes (0.5	ext{ m})}{1 	imes 10^{-3}	ext{ m}}$$
$$d = 2.945 	imes 10^{-4}	ext{ m}$$

And there we have it! By simply measuring the spread of the light on a screen, we've precisely calculated the microscopic width of the slit. This is the power of wave optics!

Visualized Solution

$S in g l e S l i tS e t u p$

$D i f f r a c t i o n P a tt er n$

$C o n d i t i o n f or M inima$

$S ma l l A n g l e A pp r o x ima t i o n$

$R e a r r an g in g t h e E q u a t i o n$

$F in d in g y$

$S u b s t i t u t i o n$

$F ina l C a l c u l a t i o n$

The Sigma Insight: Diffraction due to Single Slit

Analyzing the Setup Imagine a narrow slit of width $d$

The Master Equation The diffraction pattern on the screen consists of a broad, bright central maximum, flanked by alternating dark and bright fringes

The Small Angle Approximation In most practical setups, the wavelength of light is much smaller than the slit width, making the angle $θ$ very small

Final Calculation The problem states that the distance between the first minima on either side of the central maximum is $2 \times 1 0^{- 3} m$

Similar Questions

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Visualized Solution

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