Question

3. A dentist uses a small concave mirror of focal length 3.0 cm and holds it at a distance of 2 cm from the
tooth. What is the magnification of the image?​

Answer 1

Answer:

• magnification of image is 3.

Explanation:

Given:-

• Focal length ,f = -3 cm
• Object distance ,u = -2cm

To Find:-

• Magnification ,m

Solution:-

As we have to calculate the magnification of the image.

Firstly we calculate the image position observe by dentist.

Using Mirror Formula

• 1/f = 1/v + 1/u

where,

• v is the Image Position
• u is the object distance
• f is the focal length

Substitute the value we get

→ -1/3 = 1/v + 1/(-2)

→ -1/3 = 1/v - 1/2

→ 1/v = -1/3 + 1/2

→ 1/v = -2+3/6

→ 1/v = 1/6

→ v = 6 cm

• Hence, image position is 6 cm

Now, calculating the Magnification

• m = -v/u

Substitute the value we get

→ m = -6/-2

→ m = 6/2

→ m = 3

• Hence, the magnification of the image is 3 .

Answer 2

Answer:

Given :-

• Focal length = -3 cm
• Object Distance = -2 cm

To Find :-

Magnification

Solution :-

As we know that

1/f = 1/v + 1/u

• F is the Focal length
• V is the Image distance
• U is the object Distance

1/-3 = 1/v + 1/-2

-1/3 = 1/v - 1/2

1/v = -1/3 + 1/2

1/v = -2 + 3/6

1/v = ⅙

v = 6 × 1

v = 6 cm

Hence :-

Image distance = 6 cm

Now,

m = -v/u

m = -6/-2

m = 6/2

m = 3

Hence :-

Magnification = 3