Converging Lens Homework: Finding Image Distance with Thin Lens Equation"

[aChordate]

Homework Statement

An object is in front of a converging lens with a focal length of 11.0 cm. The lens
produces a real image located 25.0 cm from the lens. If the lens is replaced by a new lens
(in the same position) that has a focal length of 32.0 cm, what is the image distance of the
new lens? Is the image real or virtual?

Homework Equations

Thin lens equation
1/f=1/d₀+1/d_i

The Attempt at a Solution

lens 1:

1/+11*10^-2=1/d_O+1/+25.0*10^-2

d_O=0.196m

lens2:

1/+32.0*10^-2=1/0.196m+1/d_i

d_i=-0.506m

Virtual image

 

[collinsmark]

lens 1:

1/+11*10^-2=1/d_O+1/+25.0*10^-2

d_O=0.196m

lens2:

1/+32.0*10^-2=1/0.196m+1/d_i

d_i=-0.506m

Virtual image

Your approach is correct.

Your answer is pretty close, but there are some rounding errors. You might want to try again, but keep more significant figures in the intermediate steps.

 

[aChordate]

I got -.509 on the second try. Hopefully that is correct.

 

[collinsmark]

I got -.509 on the second try. Hopefully that is correct.

Yup. (Don't forget your units though.)

 

[Nickie]

I would like to commend the student for their attempt at solving the problem using the thin lens equation. However, I would like to point out that the units used in the calculation are incorrect. The focal length and image distance should both be in meters, not centimeters. So the correct equations should be:

lens 1:

1/0.11 = 1/d0 + 1/0.25

d0 = 0.0488 m

lens 2:

1/0.32 = 1/0.0488 + 1/di

di = 0.0412 m

Therefore, the image distance of the new lens is positive, indicating a real image. It is important to pay attention to units in scientific calculations as they can greatly affect the final result.