If \( I_{0} \) is the intensity of the central diffraction band in the single slit diffraction experiment, then the intensity of the band when the slit width is doubled will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 4I_{0} \)
At the centre of the single-slit pattern, the secondary wavelets from every part of the slit arrive in phase, so their amplitudes add directly. The resultant amplitude is therefore proportional to the slit width \( a \), and the intensity is proportional to the square of the amplitude.
\( \displaystyle A_{\mathrm{centre}} \propto a \;\Rightarrow\; I_{\mathrm{centre}} \propto a^{2} \)
Doubling the slit width:
\( \displaystyle \frac{I'}{I_{0}} = \left(\frac{2a}{a}\right)^{2} = 4 \;\Rightarrow\; I' = 4I_{0} \)
This agrees with the energy: twice as much light now passes through the slit, while the width of the central maximum, \( 2\lambda D/a \), is halved. Four times the intensity spread over half the width gives twice the energy.