Question

An object 15cm from cancave mirror of focal length 25cm what is the position of the image?If the object is 2cm high . what is the height of the image

Answer 1

$\large\bf\underline {To \: find:-}$

• We need to find the position of the image, and the height of the image.

$\huge\bf\underline{Solution:-}$

$\bf\underline{\red{Given:-}}$

● object distance (u) = - 15cm

● Focal length (f) = - 25cm

● size of object (h) = 2cm

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▶ By using mirror Formula :-

$\: \: \: \: \large\underline{ \boxed{ \red{ \bold{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }}}}$

➳ 1/v = 1/f - 1/u

➳ 1/v = 1/(-25) -1/(-15)

➳ 1/v = - 1/25 + 1/15

➳ 1/v = (-3 + 5)/75

➳ 1/v = 2/75

➳ 2v = 75

➳ v = 75/2

➳ v = 37.5 cm

And ,

$\red{\bf\:Magnification = h'/h = - v/u}$

⪼ h'/2 = -37.5/-15

⪼ h'/2 = 375/150

⪼ h'/2 = 25/10

⪼ h'/2 = 5/2

⪼ 2h' = 10

⪼ h' = 10/2

⪼ h' = 5cm

Hence,

• ▶ Image distance (v) = 37.5cm
• ▶ Object height (h) = 5cm

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Answer 2

We need to find the position of the image, and the height of

Solution:−

● object distance (u) = - 15cm

● Focal length (f) = - 25cm

● size of object (h) = 2cm

━━━━━━━━━━━━━━━━━━━

▶ By using mirror Formula :-

➳ 1/v = 1/f - 1/u

➳ 1/v = 1/(-25) -1/(-15)

➳ 1/v = - 1/25 + 1/15

➳ 1/v = (-3 + 5)/75

➳ 1/v = 2/75

➳ 2v = 75

➳ v = 75/2

➳ v = 37.5 cm

And ,

⪼ h'/2 = -37.5/-15

⪼ h'/2 = 375/150

⪼ h'/2 = 25/10

⪼ h'/2 = 5/2

⪼ 2h' = 10

⪼ h' = 10/2

⪼ h' = 5cm

Hence,

• Image distance (v) = 37.5cm
• Object height (h) = 5cm

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