Answer to Question #183797 in Optics for Tino

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.