Question

In Fig. 34-32, an isotropic point source of light $\mathbf{\text{S}}$is positioned at distance$\mathit{d}$from a viewing screen $\mathbf{\text{A}}$and the light intensity${\mathit{I}}_{\mathbf{P}}$at point$\mathbf{\text{P}}$(level with$\mathbf{\text{S}}$) is measured. Then a plane mirror$\mathbf{\text{M}}$is placed behind$\mathbf{\text{S}}$at distance$\mathit{d}$. By how much is${\mathit{I}}_{\mathbf{P}}$multiplied by the presence of the mirror?

Short answer

The light intensity $I_p$is multiplied by $1.11$.

Step-by-step solution

Identification of the given data

The given data is listed as follows,

• Distance between the source and the point$\text{P}$is$d$.

• The intensity of the light at point is $I_P$.

Expression of the intensity due to source

The intensity due to the source$\mathbf{\text{S}}$is expressed as follows,

${\mathbf{I}}_{\mathbf{p}}\mathbf{=}\frac{\mathbf{A}}{{\mathbf{d}}^{\mathbf{2}}}\mathbf{}$ …(i)

Here, $A$is the constant, and $d$is the distance between the source and the point.

The total intensity due to the presence of the mirror will be the sum of the intensity due to the source and the intensity due to the image.

Determination of the amount of the light intensity that is multiplied by the presence of the mirror

Now, if the plane mirror is placed behind$\text{S}$at a distance$d$, then the image will be formed at a distance$d$behind the mirror. So, the total distance from the image to the point$\text{P}$is as follows,

$d\text{'}=3d$

The intensity due to the image is as follows,

$\begin{array}{rcl}I\text{'}& =& \frac{A}{{\left(3d\right)}^2}\\ & =& \frac{A}{9d^2}\end{array}$

…(ii)

It can be observed from equations (i) and (ii),

$I\text{'}=\frac{I_p}{9}$

The total intensity at point $\text{P}$due to the presence of the mirror will be the sum of the intensity due to the source, and the intensity due to the image.

$\begin{array}{rcl}I& =& I_p+I\text{'}\\ & =& I_p+\frac{I_p}{9}\\ & =& \frac{10I_p}{9}\\ & =& 1.11I_p\end{array}$

Thus, the light intensity $I_p$is multiplied by $1.11$.