Question

A needle placed $45cm$ from a lens forms an image on a screen placed $90cm$ on the other side of the lens. What is the power of the lens?

• A. $+10D$
• B. $-10D$
• C. $+3.33D$
• D. $-3.33D$

Answer 1

The correct option is C

$+3.33D$

Step 1: Given data.

As the image forms on the other side of the lens, it is a convex lens so, the object distance is $\left(u\right)=-45cm$.

Image distance $\left(v\right)=90cm$.

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

Let the powers of the lens be $\left(P\right)$.

Step 2: Find the focal length.

Use the lens formula $\left(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\right)$.

Substitute the given values in the above formula.

$$\begin{gathered} \frac{1}{f}=\frac{1}{90cm}-\frac{1}{\left(-45cm\right)} \\ \frac{1}{f}=\frac{1}{90}+\frac{1}{45} \\ \frac{1}{f}=\frac{1}{30cm} \end{gathered}$$

So, $f=30cm$

Step 3: Convert the focal length into metres.

$\begin{array}{rcl}f& =& 30cm\times \frac{1}{100}\\ & =& 0.3m\end{array}$

Step 4: Find the power of the lens.

We know that $\left(P=\frac{1}{f}\right)$.

Substitute the given values in the above formula.

$\begin{array}{rcl}P& =& \frac{1}{f}\\ & =& \frac{1}{0.3}D\\ & =& \frac{100}{3}\\ & =& +3.33D\end{array}$

Therefore, the power of the lens is $+3.33D$.

Hence, the correct option is (C).