Question

An object of height 5 cm is placed perpendicular to the principal axis of a concave lens of focal length 10 cm. If the distance of the object from the optical centre of the lens is 20 cm, determine the position, nature and size of the image
formed using the lens formula. ​

Answer 1

Explanation:

h=5cm

focal length=-10cm

object distance= -20cm

by lens formula, we get

I÷F=I÷V+I÷U

V is -6.67 cm

Answer 2

$\large\sf{\underline{\underline{\orange{Given}}}}$

→ $\bf\green{Height\: of\: object,\: h_o = 5cm}$

→ $\bf\green{Focal\: length,\: f = -10cm}$

→ $\bf\green{Distance\: of\: object,\: u = -20cm}$

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$\large\sf{\underline{\underline{\orange{Solution}}}}$

Using lens formula,

⇛ $\bf{\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}}$

⠀

⇛ $\bf{\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}}$

⠀

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

⠀

⇛ $\bf{\dfrac{1}{v} = \dfrac{-1}{10} - \dfrac{1}{20}}$

⠀

⇛ $\bf{\dfrac{1}{v} = \dfrac{-2 -1}{20}}$

⠀

⇛ $\bf{\dfrac{1}{v} = \dfrac{-3}{20}}$

⠀

⇛ $\bf{v = -6.67}$

⠀

★ Image is at 6.67cm from concave lens.

$\bf\green{\underline{As\: magnification,}}$

⇛ $\bf{m = \dfrac{h_i}{h_o} = \dfrac{v}{u}}$

⠀

⇛ $\bf{h_i = ho × \dfrac{v}{u}}$

⠀

⇛ $\bf{h_i = 5 × \dfrac{-20}{3} × \dfrac{1}{-20}}$

⠀

⇛ $\bf{h_i = \dfrac{5}{3}}$

⠀

⇛ $\bf{h_i = 1.67cm}$

☃️ Image is virtual, erect and diminished.

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