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 solve this problem, we need to understand how the photoelectric current produced is related to the intensity of light and the focal length of the lens used. When a convex lens is used to form an image on a photo-electric cell, the intensity of light at the image location depends on the focal length of the lens. 

When the lens focuses light on the photoelectric cell, the energy incident on the cell is proportional to the intensity of the light (which in turn depends on the area over which the light is spread). The intensity \( I \) (not photoelectric current, but intensity) can be given as:

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

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

Initially, the focal length of the lens was 40 cm. When we replace the lens with another one having a focal length of 20 cm (while the diameter remains the same), the new focal length is half of the original one. Thus, the intensity increases by a factor of:

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

However, this increase in intensity leads to a proportional increase in the photoelectric current produced, according to the photoelectric effect principle. Therefore, if the initial current produced was \( I \), the new current will be:

\(= 2I\)

Hence, the photoelectric current now is 2I.

Answer 2

The intensity of light incident on the photoelectric cell is proportional to the lens’s focal length and area. Since the lens diameter is unchanged, the area remains the same. Reducing the focal length from 40 cm to 20 cm increases the intensity of light four times (since light concentration doubles due to focal length halving). Therefore, the current I will double:
So, New current = \(2I\)