Question: Q. 9. A parallel beam of light of $500 \mathrm{~nm}$ falls on a narrow slit and the resulting diffraction pattern is observed on a screen $1 \mathrm{~m}$ away. It is observed that the first minima is at a distance of $2.5 \mathrm{~mm}$ from the centre of the screen. Calculate the width of the slit.
A [O.D. I, II, III 2013]
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Solution:
Ans. The distance of the $1^{\text {st }}$ minima from the centre of the screen is,
$$ x_{1} \text { dark }=\frac{\mathrm{D} \lambda}{a} $$
where,
$\mathrm{D}=$ distance of slit from screen,
$\lambda=$ wavelength of the light ,
$a=$ width of the slit.
According to question,
$$ \begin{aligned} D & =1 \mathrm{~m} \ \lambda & =500 \mathrm{~nm} \end{aligned} $$
height of first minima $=2.5 \mathrm{~mm}$
$$ \begin{aligned} 2.5 \times 10^{-3} & =\frac{1 \times 500 \times 10^{-9}}{a} \ \Rightarrow \quad a & =2 \times 10^{-4} \mathrm{~m}=0.2 \mathrm{~mm} \end{aligned} $$
$a=0.2 \mathrm{~mm}$.
Commonly Made Errors
- Number of candidates used incorrect formula.
- Some of the students did not know the correct meaning of the symbols i.e., ’ $D$ ’ and ’ $d$ ‘, hence they interchanged them.
- Some students did not convert ’ $n \mathrm{~m}$ ’ to $\mathrm{m}$ as well as ‘mm’ to $\mathrm{m}$.
