Question

What is the size of the image produced by the converging lens in Holdin Owt’s camera if the focal length of the lens is 35 mm and the 2.0 m high object is 6.0 m away from the lens?  (Hints: First find the image distance and make sure all distances are expressed in the same units)

Answer 1

Here,

u= -6000 mm

f = + 35 cm

v= ?

Using lens formula,

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

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\ \\\\ v=\dfrac{f\times u}{f+u}=\dfrac{-210000}{-5965}=35.205\ mm$

Image will be 35.205 mm from the lens and on the opposite side as object.

Now, $\dfrac{h_i}{h_o}=\dfrac{v}{u}\\$

So, height of image $h_i=\dfrac{v\times h_o}{u}=\dfrac{35.205\times 2000}{-6000}=-11.735 \ mm$

Height of image will be 11.735 mm but it's inverted and on the other side of lens as compared to object.