The image of a candle flame placed at a distance of 36 cm from a spherical lens is formed on a screen placed at a distance of 72 cm from the lens. Identify the type of lens and calculate its focal length. If the height of the flame is 2.5 cm, find the height of the image.

Object distance $=u=-36cm$

Image distance $=v=+72cm$

Height of the object $={h}_{1}=2.5cm$

To find = height of the image $({h}_{2})$

Solution

Applying the given values on lens formula, we get-

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{f}=\frac{1}{72}-\frac{1}{(-36)}$
$\frac{1}{f}=\frac{1}{72}+\frac{1}{36}$

$\frac{1}{f}=\frac{1+2}{72}$

$\frac{1}{f}=\frac{3}{72}$

$\frac{1}{f}=\frac{1}{24}$

$f=+24cm$

Since the focal length is positive, so the lens will be a convex lens with a focal length of 24cm.

Now, magnification is given by the relation-

$m=\frac{v}{u}$

Substituting the required value in the equation we get-

$m=\frac{72}{-36}$

$m=-2$

Also, in terms of image height, magnification is given by the relation-

$m=\frac{{h}_{2}}{{h}_{1}}$

Substituting the required value we get-

$-2=\frac{{h}_{2}}{2.5}$

${h}_{2}=-2\times 2.5$

${h}_{2}=-5cm$

Hence, the height of the image will be -5cm. The negative $(-)$ represents that the image is inverted. The nature of the image formed will be, Real, Inverted and Magnified.