Question

power oflensis -10 D placed at 25cm from lens find distance or image also nature​

Answer 1

Answer :

Power of lens = -10D

Distance of object = 25cm

We have to find distance of image$.$

We can simply solve this type of questions by using lens formula. :D

• 1/v - 1/u = 1/f

Focal length of lens is not given in the question! But, Power of lens is given. :)

Relation between focal length of lens and power of lens is given by

• P = 1/f

[Remember : Unit of focal length m]

◈ Focal length of lens :

⇒ P = 1/f

⇒ f = 1/P

⇒ f = 1/(-10)

⇒ f = -0.1m = -10cm

◈ Position of image :

⇒ 1/v - 1/u = 1/f

⇒ 1/v = 1/f + 1/u

⇒ 1/v = 1/(-10) + 1/(-25)

⇒ 1/v = -1/10 - 1/25

⇒ 1/v = (-5 - 2)/50

⇒ v = -50/7

⇒ v = -7.14cm :D

Nature of image :

» Virtual and erect because lens is concave. (-ve focal length)

Answer 2

Given ,

Power of lens (P) = -10 D

Object distance (u) =- 25 cm

We know that ,

$\boxed{ \sf{Power \: of \: lens \: (P) = \frac{1}{F \: (in \: m)} }}$

Thus ,

-10 = 1/F

F = -1/10

F = -0.1 m or -10 m

$\therefore$ The focal length of concave lens is 10 m

Now , the lens formula is given by

$\boxed{ \sf{\frac{1}{F} = \frac{1}{v} - \frac{1}{u}} }$

Thus ,

-1/10 = 1/v - (-1/25)

1/v = -1/25 - 1/10

1/v = (-10 - 25)/250

1/v = -35/250

v = -250/35

v = -7.14 cm

Therefore ,

• The image distance is 17.4 cm

• The nature of image is virtual and eract