Question

An object $3.0cm$ high is placed perpendicular to the principal axis of a concave lens of focal length $15.0cm$. The image is formed at the distance of $10.0cm$ from the lens. Calculate a. the distance of object placed and b. size and nature of image formed.

Answer 1

Step 1 - Given data and assumptions made

The height of the object, $h=3.0cm$

Focal length of the concave lens, $f=-15.0cm$ (The negative sign shows that the focal length of the concave lens is negative.)

The image distance is $v=-10.0cm$ (The negative sign shows that the image is formed on the same side as that of the object in a concave lens.)

Assume the object distance as, $u$.

Assume the image height as $h\text{'}$.

Step 2 - Finding the object's distance.

The lens formula is given by, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$. Substitute the values into the formula to calculate the object distance.

$$\begin{gathered} \frac{1}{-10cm}-\frac{1}{u}=\frac{1}{-15cm} \\ \frac{1}{u}=\frac{1}{15cm}-\frac{1}{10cm} \\ \frac{1}{u}=\frac{2-3}{30cm} \\ u=-30cm \end{gathered}$$

The negative value shows that the image and the object are on the same side of the lens which means the image is virtual.

Step 3 - Finding the height and nature of the image.

It is known that the magnification in terms of object and image distance is given by, $m=\frac{v}{u}$. Substitute the values into the formula.

$$\begin{gathered} m=\frac{-10.0cm}{-30cm} \\ =+\frac{1}{3} \end{gathered}$$

As it can be seen that the magnification is positive, the image is erect.

Now, the formula of magnification in terms of the size of the image and object is given by, $m=\frac{h\text{'}}{h}$. Substitute the values into the formula.

$$\begin{gathered} \frac{1}{3}=\frac{h\text{'}}{3cm} \\ h\text{'}=1.0cm \end{gathered}$$

a. The distance of the object is $-30cm$. b. Size of the image is $1.0cm$. The image is diminished, erect, and virtual.