Question

A U-shaped wire is placed before a concave mirror having radius of curvature 20 cm as shown in figure. Find the total length of wire in the image.

• A. 10 cm
• B. 20 cm
• C. 30 cm
• D. 5 cm

Answer 1

The correct option is A 10 cm

The approach that we follow in the given situation is to get the image of a 2-D object is to locate the image of the two extreme points and finally join these images of two extreme points to get the full image of the 2-D object.

From the diagram we see that we can take two extended object and apply mirror formula for them.

Let's name them AB & CD.

First of all for AB,

u = − 30cm, f = − 10cm

We get, by applying mirror formula,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
Or,
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

$=-\frac{1}{10}--\frac{1}{30}$

$=-15cm$

So the image is at a distance 15cm from the mirror on the left side.

Now let's find the size of this image.

We have the magnification formula

$\frac{h_i}{h_o}=-\frac{v}{u}=-\frac{15}{30}=\frac{-1}{2}$.

a negative sign indicates here that the image is inverted

Or,

$h_i=\frac{-1}{2}\times 10cm=-5cm$

Or, the magnitude of the height is 5 cm.

Now, the same process we will repeat for our second extended object CD.

We have,

u = − 40 cm, f = − 10 cm

We get by applying mirror formula,

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

Or,

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

$=-\frac{1}{10}--\frac{1}{40}$

$v=-\frac{40}{3}cm$

So the image is at a distance $\frac{40}{3}cm$ from the mirror.

Now, for the size of the image, we will use the magnification formula

$\frac{h_i}{h_o}=-\frac{v}{u}=\frac{40}{3}\times \frac{-1}{10}=-\frac{1}{3}$

a negative sign indicates here that the image is inverted
Or,
$h_i=\frac{-1}{3}\times 10cm=\frac{-10}{3}cm$

Or, the magnitude of the height is $\frac{10}{3}cm.$

So, finally the image looks like this:

We see that the total length of the image is

$\frac{5}{3}cm+5cm+\frac{10}{3}cm=10cm.$