Question

A4.5 cm of length of needle placed perpendicular to the principal axis 12 cm away from the convex mirror of focal length 15cm give the magnitude of size of the image of the needle​

Answer 1

Given that,

Size of the needle, h = 4.5 cm

Object distance, u = -12 cm (negative by convention)

Focal length of the convex mirror, f = +15 cm

To find,

The magnitude of size of the image of the needle​.

Solution,

Firstly, let's find the image distance v using mirror's formula such that,

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}-\dfrac{1}{(-12)}\\\\v=6.7\ cm$

The object is placed at a distance of 6.7 cm.

Let h' is the size of the image of the needle. Using formula of magnification,

$m=\dfrac{-v}{u}=\dfrac{h'}{h}\\\\h'=\dfrac{-vh}{u}\\\\h'=\dfrac{-6.7\times 4.5}{-12}\\\\h'=2.51\ cm$

So, the height of the image of the needle is 2.51 cm.

Answer 2

Hola ⚘⚘

Answer:

• 2.51 cm

Explanation:

Given:

• Size of the needle = 4.5 cm
• Object distance = -12 cm
• Focal length of the convex mirror = +15 cm

Tofind:

• The magnitude of size of the image of the needle

Calculation:

• 1/v + 1/u = 1/f

>> 1/v = 1/f - 1/u

>> 1/u = 1/15 - 1/(-12)

>> v = 6.7 cm

• m = -v/u = h'/h

>> h' = -vh/u

>> h' = -6.7×4.5/-12

>> h' = 2.51 cm

hence, height of the image of the needle is 2.51 cm