In a Young’s double slit experiment, each slit has a width of \( 0.04\ \mathrm{mm} \) and the distance between the slits is \( 0.20\ \mathrm{mm} \). The incident light has a wavelength of \( 480\ \mathrm{nm} \) and the screen is placed at a large distance. How many total bright interference fringes are observed within the central diffraction maximum of the single slit diffraction?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 11
Note: The given answer is as per the official NTA Answer Key. However, none of the given options appears to be correct, because the bright fringes of order ±5 fall exactly on the first minima of the single-slit diffraction pattern and are therefore missing. The correct number of bright fringes within the central diffraction maximum is 9, which is not among the options, as explained below.
The interference fringes are seen inside the central diffraction maximum, which extends up to the first diffraction minima.
\( a = 0.04\,\mathrm{mm},\ d = 0.20\,\mathrm{mm} \)
First diffraction minima (small angles):
\( a\sin\theta = \pm\lambda \ \Rightarrow\ \theta_{1} = \pm\lambda/a \)
Interference maxima:
\( d\sin\theta = m\lambda \ \Rightarrow\ \theta_{m} = m\lambda/d \)
A bright fringe lies inside the central maximum when:
\( \displaystyle \frac{|m|\lambda}{d} \lt \frac{\lambda}{a} \ \Rightarrow\ |m| \lt \frac{d}{a} = \frac{0.20\,\mathrm{mm}}{0.04\,\mathrm{mm}} = 5 \)
So the bright fringes inside are \( m = -4, \ldots, 0, \ldots, +4 \), which is 9 bright fringes.
At \( m = \pm5 \), \( \theta = 5\lambda/d = \lambda/a \) exactly, where the single-slit intensity is zero, so these are missing orders.
Counting \( m = -5 \) to \( +5 \) gives \( 2\times5 + 1 = 11 \), the answer given by NTA, which seems to count the two zero-intensity fringes at the edges as well.