Question

A combination of convex and concave lens has power $4\,{\text{D}}$. If the convex lens has power $5\,{\text{D}}$, the focal length of the concave lens will be:
A. $100\,{\text{cm}}$
B. $- 100\,{\text{cm}}$
C. $- 1\,{\text{cm}}$
D. $- \dfrac{{100}}{9}\,{\text{cm}}$

Answer 1

In order to solve this numerical, use the formula of power for a combined lens system, to find the power of an individual lens. Find the focal length of that lens, which is the inverse of its power.

Complete step by step answer:
Let the combined power of the two lenses be $P$ .
Given, combined power is $4\,{\text{D}}$ .
i.e. $P = 4\,{\text{D}}$
Let the power of the convex lens be ${P_1}$ and the power of the concave lens be ${P_2}$ .
i.e. ${P_1} = 5\,{\text{D}}$
According to the formula of combined power for lenses, we have:
$P = {P_1} + {P_2}$ …… (1)
Substituting the values,
$P = 4\,{\text{D}}$ and ${P_1} = 5\,{\text{D}}$ in equation (1):
P = {P_1} + {P_2} \\\ \implies 4\,{\text{D}} = 5\,{\text{D}} + {P_2} \\\ \implies {P_2} = 4\,{\text{D}} - 5\,{\text{D}} \\\ \implies {P_2} = - 1\,{\text{D}} \\\
The formula which relates power and focal length of a lens is given by:
$P = \dfrac{1}{f}$
Where,
$P$ indicates power of the lens.
$f$ indicates the focal length of the lens, where the unit is in metres.
But here the lens is the second one, so the formula can be re-written as:
${P_2} = \dfrac{1}{{{f_2}}}$ …… (2)
Substituting the value, ${P_2} = - 1\,{\text{D}}$ in equation (2), we get:
- 1\,{\text{D}} = \dfrac{1}{{{f_2}}} \\\ \implies {f_2} = - \dfrac{1}{{1\,{\text{D}}}} \\\ \implies {f_2} = - \dfrac{1}{{1\,{{\text{m}}^{ - {\text{1}}}}}} \\\ \implies {f_2} = - 1\,{\text{m}} \\\
We know that one metre is equal to one hundred centimetres.
So,
{f_2} = - 1\, \times 100\,{\text{cm}} \\\ {\text{ = }} - {\text{100}}\,{\text{cm}}\, \\\
Hence, the focal length of the concave lens is $- {\text{100}}\,{\text{cm}}$ .

So, the correct answer is “Option B”.

Additional Information:
At the middle, a concave lens is thinner and at the edges, thicker. It is generally called a diverging lens. In the centre, a convex lens is thicker, and the edges are thinner. It is generally called converging glass. The power of a convex lens is positive, while the power of a convex lens is negative.

Note:
In this problem, you are asked to find the focal length of the concave lens in the combination. For this, find the power of the concave lens in combination. It is important to note that the power of lens corresponds to the inverse of metre. Don’t confuse it with the unit of focal length, it is in metres. You have to convert it into centimetres. Always remember that the sign of the focal length of a concave lens is negative.