Question

An electric lamp and a candle produce equal illuminance at a photometer screen when they are placed at 80 cm and 20 cm from the screen respectively. The lamp is now covered with a thin paper which transmits 49% of the luminous flux. By what distance should the lamp be moved to balance the intensities at the screen again?

Answer 1

The distance lamp be moved to balance the intensities at the screen again is 24 cm.

Given:

Distance of electric lamp = 80 cm

Distance of candle = 20 cm

To find:

Distance lamp be moved = ?

Solution:

Let I₁ be intensity of light from electric lamp.

Let I₂ be intensity of light from candle.

Now, the ratio of intensity is:

$\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{80}{20}\right)^{2}=16$

Let the distance between lamp and the screen is 'x'

The ratio is of intensity after covering the lamp is given as:

$\frac{0.491_{1}}{I_{2}}=\left(\frac{x}{20}\right)^{2}$

$0.49 \times 16 \times 400=x^{2}$

$\mathrm{x}=56 \ \mathrm{cm}$

Thus, the distance lamp moved is:

80 cm - 56 cm = 24 cm

Answer 2

At a distance of 24 cm, the lamp should be moved to balance the intensities at the screen again.

Explanation:

Let $I_1$ be the intensity at distance 80 cm and intensity at 20 cm from the plate.

Now,

$\frac{I_{1}}{I_{2}}=\left(\frac{80}{20}\right)^{2}$

     = 16

According to question "x" be the new distance between the lamp and the screen and that even after covering the lamp with thin paper, the screen intensity is balanced.

So,

$\frac{0.49 \times I_{1}}{I_{2}}=\left(\frac{x}{20}\right)^{2}$

$0.49 \times 16 \times 400=x^{2}$

$x$ = 56

Lamp has to be moved by  24 cm. i.e 80 cm - 56 cm  = 24 cm.

Thus, the lamp should be moved to a distance of 24 cm.