Question

Velocity of 2nd maxima w.r.t central maxima at $t=2s$ is

• A. $\left(8cm{s}^{-1}\right)\hat{i}+20m{s}^{-1}\hat{j}$
• B. $8cm{s}^{-1}\hat{i}$
• C. $2cm{s}^{-1}\hat{i}$
• D. $86m{s}^{-1}\hat{i}$

Answer 1

The correct option is C $2cm{s}^{-1}\hat{i}$

AS the screen will move downward with an acceleration $g=10m{s}^{-2}$

Now velocity of screen will be the velocity of central maxima.

Thus at $t=2s$, $v_c=0+g\left(5\right)=20m{s}^{-1}\hat{y}$

Position of 2nd maxima, $y=\frac{2\lambda D}{d}$

Differentiating, $v_2=\frac{2\lambda }{d}\frac{d}{dt}\left(D\right)$

As $D=1+\frac{1}{2}g{t}^2$

Thus $v_2=\frac{2\lambda gt}{d}=2cm{s}^{-1}\hat{i}$

$\therefore$ Velocity of 2nd maxima w.r.t central maxima, $v=2cm{s}^{-1}\hat{i}$