Question

An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size, and nature of the image formed.

Answer 1

Step 1; Given:

The height of the object, $h_o=5cm$

Position of the object, $u=–25cm$

The focal length of the lens, $f=10cm$

Nature of image: Real and inverted

Step 2:

Finding the position of the image:

We know that $\frac{1}{v}–\frac{1}{u}=\frac{1}{f}$

Substituting the known values in the above equation we get,

$$\begin{gathered} \frac{1}{v}+\frac{1}{25}=\frac{1}{10} \\ =>\frac{1}{v}=\frac{1}{10}–\frac{1}{25} \\ =>\frac{1}{v}=\frac{(5–2)}{50} \\ =\frac{3}{50} \\ v=\frac{50}{3}=16.66cm \end{gathered}$$

Therefore, the distance of the image is 16.66 cm on the opposite side of the lens.

Step 3: Finding the size and the nature of the image:

Now, we know that $Magnification=\frac{v}{u}$

Hence, $m=\frac{16.66}{-25}=-0.66$

Also, we know that

$m=\frac{heightofimage}{heightoftheobject}$

Or, $-0.66=\frac{heightofimage}{5cm}$

Hence, the height of the image $=-3.3cm$

The negative sign of the height of the image depicts that an inverted image is formed.

Position of image: At 16.66 cm on the opposite side of the lens

Size of image: – 3.3 cm at the opposite side of the lens