Question

An object is placed in front of a convex mirror 40 cm and of focal length 20 cm ,find the nature ,size and position of image​

Answer 1

Answer:

Since focal length of concave mirror given is 20cm l. Object is at 40cm distance which means it is at centre of curvature. hence, image will be formed of same size and at centre focus only but real and inverted.

Answer 2

Concept:-

Here, It is given the object distance from the pole and the focal length of the mirror. We have to find the nature, size and position of the image. To find this we use some certain formula. Let's Find

Given:-

• Distance of object from pole is of 40 cm.
• Focal length of the mirror is of 20 cm.

To Find:-

• Size, Nature and Position of the image ?

Assumption:-

Let, Object-distance be u.

      Image-distance be v.

      Focal-length be f.

      Image-height be $\sf{h_e}$.

      Object-height be $\sf{h_o}$.

Solution:-

Now, we use certain formula to find the Position of the Image.

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

$\sf{Here,\ u = -40\ cm\ \ and\ \ f = 20\ cm}$

So, putting the values we get,

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

$\sf{\Longrightarrow \dfrac{1}{v} = \dfrac{1}{20} - \bigg( \dfrac{1}{-40}\bigg)}$

Taking L.C.M of 20 and 40,

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

$\sf{\Longrightarrow \dfrac{1}{v} = \dfrac{3}{40}}$

By Cross multiplication we get,

$\sf{\Longrightarrow 3v = 40}$

$\sf{\Longrightarrow v = \dfrac{40}{3}\ cm\ \ or\ \ 13\dfrac{1}{3}\ cm}$        

$\sf{Hence,\ Position\ of\ image\ is\ \dfrac{40}{3}\ cm\ behind\ the\ mirror.}$

Note:- [The image will be behind the mirror because it is positive and from sign convention of convex mirror we get the positive image behind the mirror and convex mirror forms image behind the mirror.]

Now, to find size of the image we will use certain formula

$\sf{\Longrightarrow \dfrac{h_o}{h_e} = \dfrac{-v}{u}}$

$\sf{Here,\ u = 40\ cm\ \ and\ \ v = \dfrac{40}{3}\ cm}$

So, Putting all the values we get,

$\sf{\Longrightarrow \dfrac{h_o}{h_e} = \dfrac{\dfrac{-40}{3}}{-40}}$

$\sf{\Longrightarrow \dfrac{h_o}{h_e} = \dfrac{1}{3}}$

By cross multiplication we get,

$\sf{\Longrightarrow 3h_o = 1h_e}$

$\sf{\Longrightarrow h_e = 3h_o}$

$\sf{Hence, Image\ is\ 3\ times\ greater\ than\ size\ of\ object.}$

Nature $\longrightarrow$  Virtual, Erect and Diminished.

The positive sign shows the image is virtual and erect.