Answer to Question #209007 in Optics for chao

Question

compute the change in the position of the image formed by a lens with a focal length of 1.5cm as the light source is moved from its position at 6m from the lens to infinity

Answer 1

Answer:-

$\frac{1}{6 \ m}+\frac{1}{q}=\frac{1}{0.015 \ m}\\ \frac{1}{q}=\frac{1}{0.015 \ m}-\frac{1}{6 \ m}\\ \boxed{q=0.015 \ m}$

Calculate the position of the image when the light source is at infinity.

$\frac{1}{\infin}+\frac{1}{q}=\frac{1}{0.015 \ m}\\ 0+\frac{1}{q}=\frac{1}{0.015 \ m} \\ \boxed{q=0.015 \ m}$

Hence, the image is at the focus when the light source is at infinity.