QJEEM24Ph42e9: JEE Main 2024 Physics - Class 12 JEEM24Ph42e9 Solution: An effective power of a combination of 5 identical convex…

When we have a combination of identical lenses in contact, the effective power ( $P_{\text{eff}}$ ) of the combination can be calculated as the sum of the powers of all the individual lenses. This is because the lenses are in direct contact, and their powers effectively add up.

Given that the effective power of a combination of 5 identical convex lenses is $25 \mathrm{D}$ , we can use the formula for the effective power of the combination:

$P_{\text{eff}} = nP$

where:

• $P_{\text{eff}}$ is the effective power of the combination,
• $n$ is the number of lenses, and
• $P$ is the power of each individual lens.

Given $n = 5$ and $P_{\text{eff}} = 25 \mathrm{D}$ , we can solve for $P$ , the power of each lens:

$25 \mathrm{D} = 5P$

Dividing both sides by 5:

$P = \frac{25 \mathrm{D}}{5} = 5 \mathrm{D}$

The power of a lens ( $P$ ) is related to its focal length ( $f$ ) by the equation:

$P = \frac{1}{f}$

where $P$ is in diopters (D) and $f$ is in meters. Thus:

$5 \mathrm{D} = \frac{1}{f}$

Solving for $f$ gives:

$f = \frac{1}{5} \text{ meters} = \frac{1}{5} \times 100 \text{ cm} = 20 \text{ cm}$

Therefore, the focal length of each of the convex lenses is 20 cm. So, the correct answer is Option B: 20 cm.