Question

A single slit of width $a$ is illuminated by violet light of wavelength $4000\dot{A}$ and the width of the central maximum is measured as $y$. When half of the slit width is covered and illuminated by yellow light of wavelength $6000\dot{A}$, the width of the central maxima is

• A. $\frac{3y}{2}$
• B. $\frac{y}{3}$
• C. $3y$
• D. $y$

Answer 1

The correct option is C $3y$

The width of the central maximum is given by,

$y=\frac{2\lambda D}{a}$

Case $\text{(i)}$: violet light passed through single slit

$y=\frac{2D{\lambda }_{violet}}{a}..........\left(1\right)$

Case $\text{(ii)}$: yellow light passed through single slit

Let $y^{\prime }$ = Width of central maxima

$y^{\prime }=\frac{2D{\lambda }_{yellow}}{a^{\prime }}..........\left(2\right)$

Dividing $\left(2\right)$ by $\left(1\right)$

$\frac{y^{\prime }}{y}=\left(\frac{{\lambda }_{yellow}}{{\lambda }_{violet}}\right)\left(\frac{a}{a^{\prime }}\right)$

Given that, $a^{\prime }=\frac{a}{2}$

$\Rightarrow \frac{y^{\prime }}{y}=\left(\frac{6000\dot{A}}{4000\dot{A}}\right)\left(2\right)=3$

$\therefore y^{\prime }=3y$

Hence, option $\left(\text{C}\right)$ is correct.