Question

Convex lens of focal length \(f_1\) is placed in contact with a concave lens of focal length \(f_2\). For \(f_1 > f_2\) combination will behave:

• A. like a concave lens
• B. like a convex lens
• C. like a plane glass slab
• D. none of these

Hint:

Use \(\frac{1}{F}=\frac{1}{f_1}-\frac{1}{f_2}\); if \(f_1>f_2\) the result is negative, meaning diverging.

Answer 1

The Correct Option is A

Step 1: Assign signs to the focal lengths.
For a convex (converging) lens the focal length is positive: \(+f_1\). For a concave (diverging) lens it is negative: \(-f_2\) (here \(f_1, f_2\) are magnitudes).
Step 2: Write the formula for lenses in contact.
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{(-f_2)} = \frac{1}{f_1} - \frac{1}{f_2} \]
Step 3: Apply the condition \(f_1 > f_2\).
If \(f_1 > f_2\), then \(\dfrac{1}{f_1} < \dfrac{1}{f_2}\), so
\[ \frac{1}{F} = \frac{1}{f_1} - \frac{1}{f_2} < 0 \]
Step 4: Interpret the sign.
A negative combined focal length \(F\) means the combination is diverging, i.e. it behaves like a concave lens.
Only if \(f_1 = f_2\) would \(1/F = 0\) (plane slab), and if \(f_1 < f_2\) it would be convex.
\[\boxed{\text{concave lens}}\]