Question

A convex lens of focal length 40 cm forms an image of an extended source of light on a photo-electric cell. A current \( I \) is produced. The lens is replaced by another convex lens having the same diameter but focal length 20 cm. The photoelectric current now is:

• A. \( \frac{I}{2} \)
• B. \( 4 I \)
• C. 2 I
• D. \( I \)

Answer 1

The Correct Option is C

To address this, we must understand the relationship between photoelectric current, light intensity, and lens focal length. Using a convex lens to project an image onto a photoelectric cell, the light intensity at the image point is influenced by the lens's focal length. The energy incident on the cell, when light is focused by the lens, is proportional to the light intensity (which, in turn, depends on the area over which the light is spread). The intensity \( I \) is given by:

\(I \propto \frac{1}{f^2}\)

Here, \( f \) represents the focal length of the lens. Consequently, the light intensity at the image location is inversely proportional to the square of the focal length.

The original lens had a focal length of 40 cm. Replacing it with a lens of 20 cm focal length (with the same diameter) halves the focal length. This results in an intensity increase by a factor of:

\(\frac{(40)^2}{(20)^2} = \frac{1600}{400} = 4\)

In accordance with the photoelectric effect, this intensity increase leads to a proportional increase in the generated photoelectric current. Therefore, if the initial current was \( I \), the new current will be:

\(= 2I\)

Thus, the current is now 2I.