Question

An object is placed at a distance 20cm from a conves lens of focal length 10cm.find the image distance and nature of the image?

Answer 1

Answer:

Explanation:

Image distance is 20 cm and it is real and inverted.

Explanation:

It is given that,

Object distance from the convex lens, u = -20 cm

Focal length of the convex lens, f = +10 cm

If v is the position of image with respect to the mirror. It can be calculated using lens formula as :

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

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

\dfrac{1}{v}=\dfrac{1}{10}+\dfrac{1}{(-20)}

v = 20 cm

So, the image will be formed at a distance of 20 cm behind the lens. If m is the magnification of the lens,

m=\dfrac{v}{u}

m=\dfrac{20}{-20}

m = -1 (image is real and inverted)

Answer 2

Answer:

Image distance = + 20 cm & nature of image is real and inverted.

Explanation:

object distance (u) = -20 cm

focal length (f) = +10 cm

⇒ $\frac{1}{f}$ =  $\frac{1}{v}$ - $\frac{1}{u}$

⇒$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$

⇒$\frac{1}{v} = \frac{1}{10} - \frac{1}{20}$

⇒$\frac{1}{v} = \frac{2 - 1}{20}$

⇒$\frac{1}{v} = \frac{1}{20}$

∴v = 20 cm

Magnification = $\frac{v}{u}$ = $\frac{20}{-20}$ = -1

Negative value of magnification indicates that the image formed is real & inverted.