Question

An object 5 cm high is located 20 cm from a concave mirror whose focal length is 25 cm make a accurate scale diagram and hence locate the position of the image and find its size

Answer 1

Answer:

Explanation:

ho =5cm

U =-20

f =-25

v =?

hi =?

answer :

1/f = 1/v +1/u

1/-25 = 1/v + 1/-20

1/20-1/25=1/v

25-20/25*20=1/v

-5/25*20=1/v

-5v = 500

v = -100

ho =?

hi /ho =-v /u

hi /5=-(-100)/-20

hi /5=-100/20

hi /5=-5

hi =-25

Answer 2

Concept:

We need to apply the mirror formula- 1/f = 1/v + 1/u and the magnification formula- m = Hi/Ho and m = -v/u.

Given:

Height of object = 5 cm

Distance of object from a concave mirror = -20 cm

The focal length of a concave mirror = -25 cm

(negative sign is because of the reason that the focal length in the concave mirror is always negative)

Find:

We need to determine:

the location of the image

The size of the image

Solution:

We know the mirror formula,

1/f = 1/v + 1/u where f is the focal length, u is the object distance and v is the image distance

Therefore, 1/f - 1/u = 1/v

We have, f = -25 cm, u = 20cm

Therefore, 1/v = 1/-25 - 1/-20

1/v = 1/-25 + 1/20

1/v = - 4 + 5/100

1/v = 1/100

v = 100 cm

Since the image distance is a positive sign, the nature of the image will be virtual and erect.

To know the size of the image, we first need to determine its magnification

Magnification can be represented by the formula- m = Hi/Ho where Hi is the height of the image and Ho is the height of the object

Magnification can also be represented as m = -v/u

Since, L.H.S = L.H.S.

Therefore, R.H.S = R.H.S

Hi/Ho = -v/u

Hi = -v/u  × Ho

Hi = -100/-20 × 5

HI = 25 cm

Hence, the size of the image will be enlarged.

Refer to the following attachment for the ray diagram.

Thus, the distance of the image is 100 cm and its nature is virtual and erect while its height is 25 cm and its size is enlarged.

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