**CBSE NOTES CLASS 12 PHYSICS **

**CHAPTER 10 WAVE OPTICS**

**Single slit experiment**

When **single narrow slit** is illuminated by a **monochromatic source**, a broad pattern with a **central bright region** is seen. On both sides, there are alternate dark and bright regions, the intensity becoming weaker away from the centre.

Let us consider a parallel beam of light falling normally on a single slit LN of width a. The diffracted light meets a screen. The midpoint of the slit is M. MC is perpendicular to the slit.

Consider the intensity at a point P on the screen.

The slit is further divided into smaller parts, whose midpoints are M_{1}, M_{2} etc. Straight lines joining P to the different points L, M, N, etc., can be treated as parallel, making an angle θ with the normal MC.

Different parts of the wave front can be treated as secondary sources.

Since the incoming wave front is parallel to the plane of the slit, these sources are in phase.

The path difference NP – LP between the two edges of the slit will be

NP – LP = NQ = a sin θ ≈ aθ

Similarly, if two points M_{1} and M_{2} in the slit plane are separated by y, the path difference will be

M_{2}P – M_{1}P = yθ.

Equal, coherent contributions from a large number of sources, each with a different phase need to be summed up.

**Explanation for position of minima in diffraction pattern**

The minima (zero intensity) occurs at,

$$\theta =\frac{n\lambda}{a}$$

Where, n = ±1, ±2, ±3, ....

Consider the angle θ where the path difference *a*θ is λ. Then,

$$\theta \approx \frac{\lambda}{a}$$

Now, divide the slit into two equal halves LM and MN each of size $\frac{a}{2}$. For every point M_{1} in LM, there is a point M_{2} in MN such that M_{1}M_{2} = $\frac{a}{2}$. The path difference between M_{1} and M_{2} at P,

$${\mathrm{M}}_{2}\mathrm{P}\mathrm{}\u2013\mathrm{}{\mathrm{M}}_{1}\mathrm{P}\mathrm{}=\mathrm{}\frac{\mathrm{a}\mathrm{\theta}}{2}=\frac{\mathrm{\lambda}}{2}$$

That is, the contributions from M_{1} and M_{2} are 180º out of phase and cancel each other, when,

$$\theta =\frac{\lambda}{a}$$

Therefore, contributions from the two halves of the slit LM and MN, cancel each other. The intensity is also zero for,

$$\theta =\frac{n\lambda}{a}$$

where, *n *is any integer (≠ 0).

- The angular size of the central maximum increases when the slit width
*a*decreases.

**Explanation for position of maxima in diffraction pattern**

At the central point C on the screen, the angle θ is zero. All path differences are zero and hence all the parts of the slit contribute in phase. This gives maximum intensity at C. Other secondary maxima are shown at,

$$\theta =\left(n+\frac{1}{2}\right)\frac{\lambda}{a}$$

and they go on becoming weaker and weaker with increasing *n*.

Consider an angle

$$\theta =\frac{3\lambda}{2a}$$

which is midway between two of the dark fringes.

Divide the slit into three equal parts. If we take the first two thirds of the slit, the path difference between the two ends would be

$$\frac{2}{3}a=\frac{2a}{3}\times \frac{3\lambda}{2a}=\lambda $$

The first two-thirds of the slit can therefore be divided into two halves which have a path difference of $\frac{\lambda}{2}$. The contributions of these two halves cancel as described earlier. Only the remaining one-third of the slit contributes to the intensity at a point between the two minima. This will be much weaker than the central maximum, where the entire slit contributes in phase.

We can similarly show that there are maxima at

$$\theta =\left(n+\frac{1}{2}\right)\frac{\lambda}{a}$$

* *Where, *n *= 2, 3, etc.

These become weaker with increasing *n*, since only one-fifth, one-seventh, etc., of the slit contributes in these cases.