Question

an object 4cm in size, is placed at 25cm in front of a concavr mirror of focal length 15cm.at wt distance from the mirror would a screen be placed in order to obtain a sharp image? find the nature and the size of the image​

Answer 1

Given :

In concave mirror,

Object height = - 4cm.

Object distance = 25cm.

Focal length = - 15cm.

Remember !

In concave mirror focal length is negative.

To find :

(a) Image distance

(b) Size of image

(c) Nature of image

Solution :

(a) using mirror formula .i.e.,

The relationship between image distance, object distance and focal length of a spherical mirror is known as Mirror formula.

The mirror formula can be written as :

$\boxed{ \bf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }$

where,

• v denotes image distance.
• u denotes object distance.
• f denotes focal length.

substituting all the given values in the formula,

$\leadsto{ \sf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }$

$\leadsto{ \sf \dfrac{1}{v} + \dfrac{1}{( - 25)} = \dfrac{1}{ - 15} }$

$\leadsto{ \sf \dfrac{1}{v} - \dfrac{1}{25} = - \dfrac{1}{15} }$

$\leadsto{ \sf \dfrac{1}{v} = - \dfrac{1}{15} + \dfrac{1}{25} }$

$\leadsto{ \sf \dfrac{1}{v} = - \dfrac{ 7 + 3}{75} }$

$\leadsto{ \sf \dfrac{1}{v} = - \dfrac{10}{75} }$

$\leadsto{ \sf v= - \dfrac{7.5}{10} }$

$\leadsto \underline{\boxed{ \sf v = - 7.5 \: cm}}$

thus, the Image distance is -7.5cm

(b) Magnification :

The magnificent produced by a mirror is equal to the ratio of the image distance to the object distance with a minus sign,

$\leadsto{ \bf m= - \dfrac{v}{u} }$

$\leadsto{ \sf m= - \dfrac{( - 7.5)}{25} }$

$\leadsto{ \sf m= 0.3 }$

we know that,

$\leadsto{ \bf m= \dfrac{h'}{h} }$

$\leadsto{ \sf 0.3= \dfrac{h'}{4} }$

$\leadsto{ \sf h' = 0.3 \times 4 }$

$\leadsto \underline{\boxed{ \sf h' = 1.2 \: cm }}$

thus, Image height h' is 1.2cm.

(c) Nature of image :

• The image is real and inverted.
• The image is diminished.