Question

[6] An object of height 7 cm is kept at a distance of 25 cm in front of a concave mirror. The focal
length of the mirror is 15 cm. At what distance from the mirror should a sereen be kept to
clear image? What will be the size and nature of the image?​

Answer 1

Answer:

According to the question,

Object distance (u) = -25 cm;

Height of object (Ho) = + 7 cm;

Focal length (f) = -15 cm;

Image distance = v cm;

By Mirror Formula:Hi/Ho = - v/u

1/v + 1/u = 1/f.

Putting values we get,

1/v +(1/-25) = 1/ -15 cm.

⇒

1/v = 1/-15 cm - (1/-25 cm)

⇒

1/v =1/-15cm + 1/ 25cm

Taking the lcm, we get,

⇒ 1/v =-5 +3

---------

75 cm

⇒1/ v = - 2/ 75 cm

⇒v = 75/ 2 cm

⇒ v = -37.5 cm

∴ the image distance = 37.5 cm (The negative sign indicates that the image is formed on the left side of the mirror)

Now, image distance (v) = -37.5 cm

Magnification = Hi /Ho = - v/ u

Hi is the image height.

HO is the object height.

“v” is the image distance

“u” is the object distance

⇒Hi/ 7cm = (-37.5 cm/-25cm)

⇒Hi = (-37.5 cm/-25 cm) × 7 cm

⇒ Hi = -10.5 cm.

Size of image is 10.5 cm which is negative hence image is real .

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Answer 2

Solution :-

Here, a concave mirror is given

Object height ( h2) = 7

Object distance ( u) = -25

Focal length of the mirror (f) = -15

Now,

By using mirror formula,

1/f = 1/v + 1/u

1/f - 1/u = 1/v

1/v = 1/f - 1/u

Put the required values,

1/v = 1/( -15) - 1 / ( -25)

1/v = -1/15 + 1/25

1/v = -5 + 3 / 75

1/v = -2/75

v = -75/2 = -37.5

Thus, Image distance is -37.5 .

[ When the image distance is negative it means that the image is placed on the left side of the mirror ]

Now,

By using formula of magnification ,

m = h1/h2 = - v / u

h1 / h2 = - (-37.5 / -25 )

h1/7 = -1.5

h1 = - 1.5 * 7

h1 = - 10.5

Hence,The image distance and size is -37.5 and -10.5 and the image is real and inverted and placed on the left side of the mirror $.$