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## Homework Statement

An object is placed 400 mm in front of a convex lens of focal length 80 mm. Find the position of the image formed. State the nature of this image.

A second convex lens of magnifying power X8 is placed 125 mm behind the first convex lens.

What is the focal length of this lens?

Find the position of the final image formed and state its nature.

## The Attempt at a Solution

So the first part is pretty easy,

[itex]\frac{1}{U} + \frac{1}{V} = \frac{1}{F} [/itex]

[itex] \frac{1}{V} = \frac{1}{80} - \frac{1}{400} [/itex]

[itex]V = 100 mm[/itex]

Next part is where i'm unsure. Here's my attempt,

[itex]m = \frac{-V}{U}[/itex]

[itex]8 = \frac{-V}{U}[/itex]

[itex]8 = \frac{-V}{25}[/itex]

[itex]V = -200mm[/itex]

[itex]\frac{1}{U} + \frac{1}{V} = \frac{1}{F} [/itex]

[itex]\frac{1}{F} = \frac{1}{25} - \frac{1}{200}[/itex]

[itex]F = 28.6 mm[/itex]

Is this the correct method?

Thanks.