A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength 6000 Å and diffraction bands are observed on a screen 0.5 m from the slit. The distance of the third dark band from the central bright band is:

We have a single-slit diffraction problem. For a slit of width $$a$$ that is illuminated by monochromatic light of wavelength $$\lambda$$, the condition for the dark (minimum-intensity) bands on either side of the central bright band is first stated as

$$a \sin\theta = m\lambda,$$

where $$m = 1,2,3,\ldots$$ denotes the order of the dark band and $$\theta$$ is the angle the corresponding ray makes with the original direction of the beam.

Because the screen is much farther from the slit than the separation of the fringes, the angles are very small, so we use the small-angle approximations $$\sin\theta \approx \tan\theta \approx \theta.$$ If the distance from the slit to the screen is $$D$$ and the distance of the $$m^{\text{th}}$$ dark band from the central bright band measured along the screen is $$y_m,$$ then $$\tan\theta = y_m/D.$$ Substituting $$\tan\theta$$ for $$\sin\theta$$ in the condition gives

$$a\,\frac{y_m}{D} = m\lambda.$$

Solving for $$y_m$$ we get the working formula

$$y_m = \frac{m\lambda D}{a}.$$

Now we insert the numerical data step by step. The width of the slit is given as $$0.1\ \text{mm}$$, so

$$a = 0.1\ \text{mm} = 0.1 \times 10^{-3}\ \text{m} = 1.0 \times 10^{-4}\ \text{m}.$$

The wavelength of light is $$6000\ \unicode{x212B}$$; recalling that $$1\ \unicode{x212B} = 10^{-10}\ \text{m},$$ we write

$$\lambda = 6000 \times 10^{-10}\ \text{m} = 6.0 \times 10^{-7}\ \text{m}.$$

The screen distance is given as $$D = 0.5\ \text{m}.$$ We need the third dark band, so $$m = 3.$$ Substituting all these values into the expression for $$y_m$$ gives

$$\begin{aligned} y_3 &= \frac{m\lambda D}{a} \\ &= \frac{3 \,\bigl(6.0 \times 10^{-7}\ \text{m}\bigr)\,(0.5\ \text{m})}{1.0 \times 10^{-4}\ \text{m}}. \end{aligned}$$

First multiply the wavelength by the screen distance:

$$6.0 \times 10^{-7}\ \text{m} \times 0.5 = 3.0 \times 10^{-7}\ \text{m}.$$

Next multiply by the order $$m = 3$$:

$$3 \times 3.0 \times 10^{-7}\ \text{m} = 9.0 \times 10^{-7}\ \text{m}.$$

Now divide by the slit width:

$$\frac{9.0 \times 10^{-7}\ \text{m}}{1.0 \times 10^{-4}\ \text{m}} = 9.0 \times 10^{-3}\ \text{m}.$$

Finally convert metres to millimetres, remembering that $$1\ \text{m} = 1000\ \text{mm}:$$

$$9.0 \times 10^{-3}\ \text{m} = 9.0 \times 10^{-3} \times 1000\ \text{mm} = 9.0\ \text{mm}.$$

So, the third dark band lies $$9\ \text{mm}$$ away from the central bright band.

Hence, the correct answer is Option A.