Question

An object is placed at a distance of X from a convex lens, when a real image is formed at a distance of$15cm$ from the lens. If the focal length of the lens is$10cm$, calculate (a) value of X (b) the magnification produced. State the characteristics of the image.

Answer 1

Step 1: Given data

Object distance from the lens$\left(u\right)=X$

Image distance from the lens$\left(v\right)=+15cm$

The focal length of the lens$\left(f\right)=+10cm$ (the lens is convex)

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Step 2: Calculating the distance of the object

Applying the lens formula to find the distance of the object,

$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{15}-\frac{1}{X}=\frac{1}{10} \\ \Rightarrow -\frac{1}{X}=\frac{1}{10}-\frac{1}{15}=\frac{3-2}{30}=\frac{1}{30} \end{gathered}$$

We get, $X=-30cm$

A negative sign shows that the distance is measured opposite to the direction of the incident light.

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Step 3: calculating the magnification

Magnification$=\frac{v}{u}=\frac{15}{-30}=-0.5$

The magnification is negative and less than one so, the image will be real, inverted, and diminished.

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The value of X is$-30cm$ and magnification is $-0.5$. The image will be real, inverted, and diminished.