Question

an object of height 4 cm is placed at a distance of 20 cm from a concave lens of focal length 10 cm use lens formula to determine the position of the image formed and height of the image​

Answer 1

$\huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

Height of object (h) = 4 cm

Object Distance (u) = 20 cm

focal length (f) = 10 cm

If Lens is concave then,

Take focal length as negative.

Use lens formula

$\Large{\boxed{\boxed{\sf{\dfrac{1}{f} \: = \: \dfrac{1}{v} \: - \: \dfrac{1}{u}}}}}$

Put Values

⇒1/v = 1/f + 1/u

Put values

⇒1/v = (1/-10) + (1/-20)

⇒1/v = -1/10 - 1/20

⇒1/v = -2 - 1/20

⇒1/v = -3/20

v = 20/3

∴ v = 6.67 cm

_______________________________

As we know that

magnification = v/u = h'/h

⇒v/u = h'/h

⇒6.67/20 = h'/h

⇒h'= 6.67*4/20

⇒h' = 6.67/5

⇒h' = 1.334

∴ Size Of Object is 1.334 cm

Answer 2

Explanation:

Hey mate..

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Given,

Size ( height ) of the object , h = 4 cm

Object Distance , u = - 20cm

Image Distance , v = ?

Focal length of a concave lens , f = -10cm

From Lens Formula we have,

======================

So,

The image is virtual and erect and it is formed at the distance of 6.67 m on the same side of the object.

Hope it helps !!