Why Is the Virtual Image in a Convex Mirror Always Smaller?

[IniquiTrance]

Homework Statement

Prove that the virtual image in a convex mirror is always smaller than the real object.

Homework Equations

m = -\frac{d_{i}}{d_{O}}

The Attempt at a Solution

Not a homework problem. Something which is bothering me, and haven't been able to prove yet.

Thanks!

 

[rl.bhat]

When the object is at infinity, where the image is formed in convex mirror. Now write the relation between di, do and f with proper sign. Multiply by di to each term on both the side and find the relation for m. From the result, see whether you get your answer.

 

[IniquiTrance]

Thanks for the response.

When the object is at infinity, the image is formed at the focus.

When I multiply both sides by d_i, I get -m + 1 = \frac{d_{i}}{f}

I still don't see why this proves |m|< 1 :(

 

[IniquiTrance]

Ah ok I see it.

m = 1 - \frac{d_{i}}{f}

Since m = -\frac{d_{i}}{d_{O}}> 0 for convex mirrors, since the image is behind the mirror, while the object is in front, m is at max 1.

Thanks!