Question

An object is place before a convex mirror of focal length 20cm at a distance of 40cm findtheposition and nature of image​

Answer 1

Given :-

After sign convention

• focal length,f =-20cm

• u = -40cm

To find :-

The position of image nature and size of the image.

${\red{\boxed{\large{\bold{Mirror\:Formula}}}}}$

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

Solution:-

By using Mirror Formula

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

$\sf\implies\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

★ Putting the values in the above formula,we get:

$\sf\implies\dfrac{1}{v}=\dfrac{-1}{20}-\dfrac{(-40)}{6}$

$\sf\implies\dfrac{1}{v}=\dfrac{-1}{20}+\dfrac{1}{40}$

$\sf\implies\dfrac{1}{v}=\dfrac{-2+1}{40}$

$\sf\implies\:v=-40cm$

since, object is placed between P and F , we get virtual and erect.

We know the magnification :

$\sf\:m=\dfrac{-v}{u}$

$\sf\implies\:m=-\dfrac{(-40)}{(-40)}$]

$\sf\:m=-1$

Now ,we assume that height of object is x cm.

We know that

$\sf\:m=\dfrac{h_{i}}{h_{o}}$

$\sf\implies\:-1=\dfrac{h_{i}}{x}$

$\sf\implies\:h_{i}=-x\:cm$

It means , image is inverted

Conclusion:-

Postion of image ,v =-40cm

Nature of image : virtual and erect

Height of image ,h = -x =(-ve)×height of object