Question

a circular plate of radius r is placed on the principal Axis at a distance from a convex mirror of focal length f such that the principal Axis passes through the centre of plate normally the area of image of the plate is​

Answer 1

The correct answer is that the area of the image would be one fourth the size of the object.

Explanation:

• The formula of the mirror is 1v + 1u = 1f
• Here the terms u,v and f represent the distances of the object as well as that of the image from the lens. This includes focal length too.
• Hence by solving the formula we get v = uf/u-f.
• Where as the magnification of the lens is f/f−u.
• We know that convex mirrors have u in negative values while f and v have positive values.
• According to this we perceive the area of the image i.e plate would be $L^{2}$/4

Answer 2

A circular plate of radius r is placed on the principal Axis at a distance f from a convex mirror of focal length f such that the principal Axis passes through the centre of plate normally. The area of image of the plate is​

πr²/4

Explanation:

Distance of the object u = -f

Focal length of the convex lens = f

Using the mirror formula

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}-\frac{1}{f}=\frac{1}{f}$

$\implies \frac{1}{v}=\frac{1}{f}+\frac{1}{f}$

$\implies \frac{1}{v}=\frac{2}{f}$

Magnification

$M=-\frac{v}{u}$

$\implies M=-\frac{f/2}{-f}$

$\implies M=\frac{1}{2}$

Thus, the radius r of the circular plate will be magnified by this amount for its image

The radius of the image will be

$r'=r\times\frac{1}{2}=\frac{r}{2}$

Thus, the area of the image of the plate will be

$A'=\pi r'^2$

$\implies A'=\pi\times(\frac{r}{2})^2$

$\implies A'=\pi\frac{r^2}{4}$

Hope this answer is helpful.

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