Question

The focal length of a convex mirror is 14 m, the distance of the object from the mirror is 8 m, the length of the object is 3 m, find the magnificent of the image and the length of the image.

No violation please

Answer 1

$\huge{\underline{\mathfrak {\purple{Answer }}}}$

It is stated that,

The focal length of the mirror(f) = 14 m

The distance of the object (u) = 8 m

The length of the object (o) = 3 m

Let, the distance of the formed image be, 'v'

The length of the formed image be, 'i'

The magnificent of the image be, 'm'

Now let's come to the calculations_

We know that,

1/focal length = (1/distance of the object) + (1/distance of the formed image)

⇒ 1/f = (1/u) + (1/v)

⇒ 1/14 = (1/8) + (1/v)

⇒ (1/14)-(1/8) = 1/v

⇒ 1/v = (1/14)-(1/8)

⇒ 1/v = (4-7)/56 [∵ The L.C.M. of 14 and 8 is 56]

⇒ 1/v = -3/56

⇒ v = -56/3

⇒ v = -18.66 m

So, the image will form at a distance of_

(-18.66+8)

= -10.66 m beyond the centre of curvature.

Magnificent of the image(m) = - v/u

                                                = - (-18.66/8)

                                                = 18.66/8

                                                = 2.3325

                                                = 2.333

So, the magnificent of the image is 2.333

We know that,

m = i/o

⇒ 2.333 = i/3

⇒ 2.333×3 = i

⇒ i =6. 999

⇒ i = 7 m

So, the length of the formed image is 7 m.

Answer 2

Answer:

The focal length of the mirror(f) = 14 m

The distance of the object (u) = 8 m

The length of the object (o) = 3 m

Let, the distance of the formed image be, 'v'

The length of the formed image be, 'i'

The magnificent of the image be, 'm'

Now let's come to the calculations_

We know that,

1/focal length = (1/distance of the object) + (1/distance of the formed image)

⇒ 1/f = (1/u) + (1/v)

⇒ 1/14 = (1/8) + (1/v)

⇒ (1/14)-(1/8) = 1/v

⇒ 1/v = (1/14)-(1/8)

⇒ 1/v = (4-7)/56 [∵ The L.C.M. of 14 and 8 is 56]

⇒ 1/v = -3/56

⇒ v = -56/3

⇒ v = -18.66 m

So, the image will form at a distance of_

(-18.66+8)

= -10.66 m beyond the centre of curvature.

Magnificent of the image(m) = - v/u

                                                = - (-18.66/8)

                                                = 18.66/8

                                                = 2.3325

                                                = 2.333

So, the magnificent of the image is 2.333

We know that,

m = i/o

⇒ 2.333 = i/3

⇒ 2.333×3 = i

⇒ i =6. 999

⇒ i = 7 m

So, the length of the formed image is 7 m

Explanation: