A 5 watt source emits monochromatic light of wavelength $5000 Å$. When placed $0.5 \mathrm{~m}$ away, it liberates photoclectrons from a photosensitive metallic surface. When the source is moved to a distance of $1.0 \mathrm{~m}$, the number of photoelectrons liberated will be reduced by a factor…

Question

A 5 watt source emits monochromatic light of wavelength $5000 Å$. When placed $0.5 \mathrm{~m}$ away, it liberates photoclectrons from a photosensitive metallic surface. When the source is moved to a distance of $1.0 \mathrm{~m}$, the number of photoelectrons liberated will be reduced by a factor of :

Aicharya · Accepted Answer

Intensity is given by
$I=\frac{p}{4 \pi d^2}$
where $p=$ power of source
$d=$ distance
We know (assume) that light to spread out uniformly is all direction i.e. spherical source.
$\begin{aligned}
& \therefore I \propto \frac{1}{d^2} \text { or } \frac{I_1}{I_2}=\frac{d_2^2}{d_1^2} \\
& \text {or } \frac{I_1}{I_2}=\left(\frac{1}{0.5}\right)^2 \\
& \Rightarrow \frac{I_1}{I_2}=4 \\
& \text {or } I_2=\frac{I_1}{4}
\end{aligned}$

A 5 watt source emits monochromatic light of wavelength $5000 Å$. When placed $0.5 \mathrm{~m}$ away, it…

Solution