Question

A person suffering from myopia (near-sightedness) was advised to wear a corrective lens of power - 2·5 D. A spherical lens of the same focal length was taken in the laboratory. Draw a ray diagram to show the position and nature of the image formed at a distance of 10 cm from the lens.

Answer 1

Nature of image formed-

1. The power of the given lens was negative, hence a diverging/concave lens was used
2. The image formed has a position on the left side of the lens same as that of the object
3. The image formed will be virtual and erect and diminished

Calculation of focal length of the lens,

Step 1: Given that

$P=-2.5D$

Step 2: Formula used

$Poweroflens\left(P\right)=\frac{1}{Focallenght\left(f\right)}$

Step 3: Calculation

$Focallength\left(f\right)=\frac{1}{P}=\frac{1}{-2.5D}=-0.4m=-40cm$

$f=-0.4m$

Calculation of position of the object:

Step 1: Given data

$u=-10cm=-0.1m$

Step 2: Formula used

The lens formula$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ , here $u$ is the distance of image, $v$ is the distance of the object and $f$ is the focal length.

Step 3: Calculation:

$\frac{1}{-0.1}-\frac{1}{u}=\frac{1}{-0.4}$

$$\begin{gathered} \frac{1}{u}=\frac{1}{0.4}-\frac{1}{0.1} \\ \frac{1}{u}=2.5-10 \\ u=\frac{1}{-7.5} \\ u=-0.13333m=-13.33cm \end{gathered}$$

The position of the object is 13.33 cm on the left and the focal length is 40 cm.

Ray diagram of the position and nature of the image formed at a distance of 10 cm from the lens:

The ray diagram given below shows the object and image.