Question

A convex lens of focal length $40cm$, a concave lens of focal length $40cm$ and a concave lens of focal length $15cm$ are placed in contact. The power of the combination in dioptre is

• A. $+1.5$
• B. $-1.5$
• C. $+6.67$
• D. $-6.67$

Answer 1

The correct option is D

$-6.67$

Step 1: Given data.

The focal length of the convex lens is $\left(f_1\right)=40cm$.

The focal length of the first concave lens is $\left(f_2\right)=-40cm$.

The focal length of the second concave lens is $\left(f_3\right)=-15cm$.

Let the total focal length be $\left(F\right)$.

Let the power be $\left(P\right)$

Step 2: Find the total focal length of the lens.

So, $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}$.

Substitute the given values in the above relation.

$$\begin{gathered} \frac{1}{F}=\frac{1}{40cm}+\frac{1}{-40cm}+\frac{1}{-15cm} \\ \frac{1}{F}=\frac{1}{40cm}-\frac{1}{40cm}-\frac{1}{15cm} \\ \frac{1}{F}=-\frac{1}{15cm} \end{gathered}$$

Step 3: Convert the total focal length into metres.

$$\begin{gathered} \frac{1}{F}=-\frac{1}{15cm} \\ \Rightarrow -\frac{1}{15cm}=-\frac{100}{15m} \end{gathered}$$

Step 4: Find the power.

As we know, $\left(P=\frac{1}{F}\right)$.

Substitute the value of $\left(\frac{1}{F}\right)$ in the above relation.

$\begin{array}{rcl}P& =& -\frac{100}{15}D\\ & =& -\frac{20}{3}D\\ & =& -6.67D\end{array}$

So, the power of the lens is $-6.67$.

Therefore, the correct option is (D).