Question

an object 4 cm high is placed 14.0 CM in the front of concave mirror of focal length 20 CM find the distance from the mirror at which screen is placed in order to obtained a Sharp image. also find the nature of the image formed​

Answer 1

Given :

• Height of the object (hi) = 4 cm.

• Object Distance (u) = 14 cm.

• Focal length (f) = 20 cm.

To find :

• Image distance (v).

• Nature of the image.

Solution :

To find the image distance :

We know the mirror formula i.e,

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

Where :

• f = Focal length
• v = Image distance
• u = Object Distance

By sign convention , we know that the focal length and object Distance is always negative in a concave mirror , so we get :

• Focal length (f) = (-20) cm

• Object Distance (u) = (-14) cm.

Now by using the mirror formula and substituting the values in it, we get :

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

$:\implies \bf{\dfrac{1}{(-20)} = \dfrac{1}{v} + \dfrac{1}{(-14)}}$

$:\implies \bf{\dfrac{1}{(-20)} - \dfrac{1}{(-14)} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{1}{(-20)} + \dfrac{1}{14} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{-7 + 10}{140} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{3}{140} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{140}{3} = v}$

$:\implies \bf{46.67(approx.) = v}$

$\boxed{\therefore \bf{v = 47\:cm}}$

Hence the Image distance is 47 cm.

Nature of the image :

We know the formula for magnification i.e,

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

Where :

• m = magnification
• v = Image distance
• u = Object Distance

Now using the formula magnification and substituting the values in it, we get :

$:\implies \bf{m = -\dfrac{v}{u}}$

$:\implies \bf{m = -\dfrac{(-47)}{(-14)}}$

$:\implies \bf{m = -\dfrac{(\not{-}47)}{(\not{-}14)}}$

$:\implies \bf{m = -\dfrac{47}{14}}$

$:\implies \bf{m = - 3.35}$

$\boxed{\therefore \bf{m = - 3.4}}$

Hence the magnification of the object is - 3.4.

Thus the image is real and inverted .