Question

An object 6cm in size is placed at 30 in front of a concave mirror of focal length 15cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find nature and size of the image?

Answer 1

v= -50cm

By sign convention:

Object Distance,u: -30cm

Focal length,f:-15cm

Image distance,v:?

Applying Mirror formula,

$$\begin{lgathered}\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \\ \\ \implies \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \\ \\ \\ \implies \: \frac{1}{v} = \frac{1}{ - 25} - \frac{1}{ - 50} \\ \\ \\ \implies \: \frac{1}{v} = \frac{ - 1}{25} + \frac{1}{50} = \frac{ - 2 + 1}{50} \\ \\ \\ \implies \: \frac{1}{v} = \frac{ - 1}{50} \\ \\ \\ \implies \: \boxed{v = - 50cm}\end{lgathered}$$

Characteristics of image formed:

1)Real and inverted

2)Same size as the object

3)Formed in front of the mirror

Now,

•Magnification factor,m:

$$m = \frac{ - v}{u} = - \frac{ - 50}{ - 50} = - 1$$

Height of object:6cm

Height of image:?

Also,

m= h`/h

=> -1=h`/6

=>h`= -6

Height of the image is -6

Answer 2

$\Large\bf {\orange {\underline { \underline {Given}} : - }}$

➝ Object distance, u = -30cm

➝ Focal length, F = -15cm

➝ Object height, h = 6cm

$\Large\bf {\orange {\underline { \underline {To \: find}} : - }}$

➝ Nature and size of the image

$\Large\bf {\orange {\underline { \underline {Solution}} : - }}$

using Mirror formula,

$\longrightarrow \sf \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \\ \longrightarrow \sf \frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

$\longrightarrow \sf \frac{1}{v} = \frac{1}{ - 25} - \frac{1}{ - 50}$

$\longrightarrow \sf \frac{1}{v} = - \frac{1}{ 25} + \frac{1}{50}$

$\longrightarrow \sf \frac{1}{v} = \frac{ - 2 + 1}{50}$

$\longrightarrow \sf \frac{1}{v} = \frac{1}{50}$

$\longrightarrow {\sf v = - 50cm}$

Hence, image distance is -50cm

Nature of image.....

• Image is real and inverted
• Same size as object
• Image will be formed infront of the mirror

$\sf magnification ,\: m = \frac{ - v}{u}$

$\longrightarrow \sf m = - \cancel \frac{ - 50}{ - 50} = - 1$

We know,

→ Object height, h = 6cm

→ Image Height, h' = ?

also,

$\sf magnification ,\: m = \frac{h'}{h}$

$\longrightarrow \sf - 1 = \frac{h'}{6}$

$\longrightarrow \sf h' = - 1 \times 6 = - 6cm$

Thus, the height of the image is -6cm