Question

The focal lengths of a convergent and a divergent lens are \( f_1 \) and \( f_2 \) respectively. If the lenses are placed in contact, then in the following conditions of combination, write the nature of the combined lens and also draw the ray diagram: (i) \( f_1>f_2 \)
(ii) \( f_1<f_2 \)
(iii) \( f_1 = f_2 \)
(i) Condition: \( f_1>f_2 \)

Hint:

The focal length of a lens combination is determined by the sum of reciprocals of individual focal lengths.

Answer 1

Step 1: The effective focal length of the lens combination is given by:

\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]

Step 2: Since \( f_1>f_2 \) and \( f_2 \) is negative for a diverging lens:

\[ \frac{1}{F} = \frac{1}{f_1} - \frac{1}{|f_2|} \]

Step 3: If \( f_1 \) dominates, the net effect is converging.

\( \text{Combination acts as a converging lens.} \)

(ii) Condition: \( f_1<f_2 \)

Solution:

Step 1: Using the same formula:

\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]

Step 2: If \( f_1<f_2 \), the negative focal length of the diverging lens dominates.

\[ \frac{1}{F}<0 \]

Step 3: The net effect is a diverging lens.

\( \text{Combination acts as a diverging lens.} \)

(iii) Condition: \( f_1 = f_2 \)

Solution:

Step 1: Using the lens formula:

\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]

Step 2: If \( f_1 \) and \( f_2 \) are equal in magnitude but opposite in sign:

\[ \frac{1}{F} = 0 \Rightarrow F = \infty \]

Step 3: The combination acts as a plane parallel glass.

\( \text{Combination acts as a plane glass.} \)