Question

an object of 5cm is placed at a distance of 16cm from a convex lense of focal length 10cm .find the nature losition and size of the image
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Answer 1

Given

• Object Distance = -16 cm
• Object Size = 5 cm
• Lens Used = Convex
• Focal length = 10 cm

To Find

• Nature, Position and size of the image

Solution

● Lens Formula ; 1/v - 1/u = 1/f

● Magnification = hᵢ/hₒ = -v/u

✭ Image Distance :

→ 1/v + 1/u = 1/f

→ 1/v + 1/-16 = 1/10

→ 1/v - 1/16 = 1/10

→ 1/v = 1/10 + 1/16

→ 1/v = 8/80 + 5/80

→ 1/v = (8+5)/80

→ 1/v = 13/80

→ v = 80/13

→ v = 6.1 cm

✭ Magnification :

→ M = -v/u

→ M = -6.1/-16

→ M = 0.3

✭ Image Size :

→ hᵢ/hₒ = -v/u

→ hᵢ = 0.3 × 5

→ hᵢ = 1.5 cm

∴ The image is virtual, erect and diminished with a size of 1.5 cm placed at 6.1 cm

Answer 2

• Given

• Height of object ($\sf{h_o}$) = 5 cm
• Object distance (u) = - 16 cm
• Focal length (f) = 10 cm

• To find

• Nature, position and size of the image.

• Solution

Distance of image from pole (v) -

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

$\implies\sf{ \dfrac{1}{v} + \bigg( - \dfrac{1}{16} \bigg) = \dfrac{1}{10} }$

$\implies\sf{ \dfrac{1}{v} - \dfrac{1}{16} = \dfrac{1}{10} }$

$\implies\sf{ \dfrac{1}{v} = \dfrac{1}{10} + \dfrac{1}{16} }$

$\implies\sf{ \dfrac{1}{v} = \dfrac{8 + 5}{80}}$

$\implies\sf{ \dfrac{1}{v} = \dfrac{13}{80}}$

$\implies\sf{ v = \dfrac{80}{13}}$

$\implies\sf{ v = \cancel{\dfrac{80}{13}}}$

$\implies\sf{v = 6.1}$

$\red{\bigstar}$$\underline{\boxed{\sf{v = 6.1~cm}}}$

Magnification -

$\underline{\boxed{\sf{m = \dfrac{-v}{u}}}}$

$\implies \sf{ \dfrac{ - 80}{13 \times - 16}}$

$\implies \sf{ \dfrac{ 80}{13 \times 16}}$

$\implies \sf{m = 0.3}$

$\red{\bigstar}$$\underline{\boxed{\sf{m = 0.3}}}$

Height of image -

$\underline{\boxed{\sf{\dfrac{h_i}{h_o} = \dfrac{-v}{u}}}}$

$\implies \sf{\dfrac{h_i}{5} = 0.3}$

$\implies \sf{h_i = 0.3 \times 5}$

$\implies \sf{h_i = 1.5~cm}$

$\red{\bigstar}$ $\underline{\boxed{\sf{h_i = 1.5~cm}}}$

•°• Image is virtual, erect and diminished.