Question

An object is placed at a distance of 40 cm from a concave lens of focal length 20cm • Find the distance of image from the lens. List the characteristics of image formed by lens nature, position, and size) • Draw ray diagram to justify your answer.​

Answer 1

★ Figure refer to attachment

Given

An object is placed at a distance of 40 cm from a concave lens of focal length 20cm • Find the distance of image from the lens.

To find

List the characteristics of image formed by lens nature, position, and size) • Draw ray diagram to justify your answer.

Solution

• Object distance (u) = -40cm
• Focal length (f) = - 20cm
• Image distance (v) = ?

Appy lens formula

$\implies\sf \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f} \\ \\ \\ \implies\sf \dfrac{1}{v}-\dfrac{(-1)}{40}=\dfrac{-1}{20} \\ \\ \\ \implies\sf \dfrac{1}{v}+\dfrac{1}{40}=\dfrac{-1}{20} \\ \\ \\ \implies\sf \dfrac{1}{v}=\dfrac{-1}{40}-\dfrac{1}{20} \\ \\ \\ \implies\sf \dfrac{1}{v}=\dfrac{-1-2}{40} \\ \\ \\ \implies\sf \dfrac{1}{v}=\dfrac{-3}{40} \\ \\ \\ \implies\sf v=\cancel\dfrac{-40}{3} \\ \\ \\ \implies\sf v=-13.3cm$

Hence, the image distance is -13.3cm

Nature of image

• Nature = Virtual and erect
• Size = Diminished
• Position = b/w F1 and optical centre

Answer 2

• ⇬⇬ Ray Diagram ⇬⇬

Gɪᴠᴇɴ :-

ᶜᵒⁿᶜᵃᵛᵉ ᴸᵉⁿˢ

• Object distance (u) = -40 cm
• Focal length (f) = -20 cm

ᴛᴏ ғɪɴᴅ :-

• Image distance (v)
• Nature, Size, Position of Image

sᴏʟᴜᴛɪᴏɴ :-

On using Lens Formula, we get,

$\Large{\boxed{\frac{1}{f}=\frac{1}{v}-\frac{1}{u}}}$

where,

f = focal length

v = image distance

u = object distance

On putting above given values in it, we get,

$\implies \: \frac{-1}{ 20} = \frac{1}{v} - \frac{-1}{ 40} \\ \\ \\ \implies \: \frac{ - 1}{20} = \frac{1}{v} + \frac{1}{40} \\ \\ \implies \: \frac{1}{v} = \frac{ - 1}{20} - \frac{1}{40} \\ \\ \\ \implies \: \frac{1}{v} = \frac{ - 1 \times 2}{20 \times 2} - \frac{1}{40} \\ \\ \\ \implies \: \frac{1}{v} = \frac{ - 2 - 1}{40} \\ \\ \\ \implies \: \frac{1}{v} = \frac{ - 3}{40} \\ \\ \\ \implies \: v = \frac{ -40}{ 3} \\ \\ \\ \implies \: v = -13.33cm$

Hence,

Image distance (v) = -13.33 cm

Now,

Now, We know that,

$\implies \: Magnification(m) = \frac{Image \: distance(v)}{Object \: distance(u)} \\ \\ \implies \: m = \frac{v}{u}$

On putting above values in it, we get,

$\implies \: m = \frac{ \frac{ - 40}{3} }{ - 40} \\ \\ \implies \: m \: = \frac{ - 40}{3} \times \frac{1}{ - 40} \\ \\ \implies m = \frac{1}{3}$

Hence,

m is positive so,

• Image formed is virtual and erect and on the same side of object.
• Smaller than object
• Between F1 & optical centre