$h v=h v_{0}+K E_{\max }$
$\frac{h c}{\lambda_{\text {incident }}}=\frac{h c}{\lambda_{0}}+ eV _{ s }$
When $\lambda_{\text {incident }}=\lambda,$ then $V_{s}=3 V$
$\frac{h c}{\lambda}=\frac{h c}{\lambda_{0}}+3 eV \ldots .$ $(i)$
And for $\lambda_{\text {incident }}=2 \lambda,$ then $V_{z}=1 V$
$\frac{h c}{\lambda}=\frac{h c}{\lambda_{0}}+1 eV \ldots$ $(ii)$
On solving equations $(i)$ and $(ii)$ we get,
$\frac{2 h c}{\lambda_{0}}=\frac{1}{2} \frac{h c}{\lambda}$
$\lambda_{0}=4 \lambda$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| S.No. | $R$ | $l$ | $100-l$ | $S = \left( {\frac{{100 - l}}{l}} \right)R$ |
| $1$ | $20\,\Omega $ | $43$ | $57$ | $26.51\,\Omega $ |
| $2$ | $30\,\Omega $ | $51$ | $49$ | $28.82\,\Omega $ |
| $3$ | $40\,\Omega $ | $59$ | $41$ | $27.80\,\Omega $ |
| $4$ | $60\,\Omega $ | $70$ | $30$ | $25.71\,\Omega $ |
[Figure shows position of reference $'O'$ when jaws of screw gauge are closed]
Given pitch $=0.1 \,{cm}$.