Question

An object of height 6 cm placed at a distance of 24 cm from a concave mirror of focal length 30 cm. Find
position, height and nature of image.

please explain your answer and do not change the values and I will mark you brainliest ​

Answer 1

Answer:

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sorry I don't know this answer

Step-by-step explanation:

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Answer 2

Answer:

Correct answer would be: position of image = 120 cm, height of image = 30 cm, nature of image = Virtual, Erect and Enlarged.

Step-by-step explanation:

Given: Height of object = 6 cm

           Distance of object(u) = 24 cm from the mirror

           Focal Length of concave mirror(f) = 30 cm.

To find: Position(v) , Nature and Height of image.

Now, the object's height, distance from mirror(u) and focal length(f) is given. Apart from it, we are also told that the mirror is a concave mirror.

As we know, for a concave mirror, value of u and f are always negative. Hence, u = -24 cm and f = -30 cm.

Now, we know from the formula for spherical mirror:

1/f = 1/u+1/v

where, v is the position of image.

Hence, putting values given in the question with the applied signs:

-1/30 = -1/24 + 1/v

1/v = -1/30 + 1/24

1/v = (5-4)/120

1/v = 1/120

Therefore, v = 120 cm.

As the value of v is positive, it proves that image is Virtual and Erect.

Now, for magnification, we know the formula:

Magnification = -v/u = height of image / height of object

Hence, putting values:

(-120/-24) = Hₓ / 6

where, Hₓ = height of image. Hence,

5 = Hₓ / 6

Hₓ = 30 cm

Therefore, Height of image is 30 cm.