Question

An object is 6cm high in placed at a distance of 24cm from a concave mirror which produces a real image 5cm high.

(i) Find the position of the image. (ii) What is the focal length of the mirror.
Answer ASAP please​

Answer 1

We know that,

v

1

+

u

1

=

f

1

u = -15 cm

Focal length = -10 cm

∴

v

1

=

−10

1

−

(−15)

1

⇒

v

1

=

−10

1

+

15

1

=

30

−3+2

=−

30

1

⇒v=−30cm

Magnification (m)=

u

v

=

Object height

Image height

m=−

u

v

=+

−15

30

=−2

Also, Image height=−2×6=−12 cm

Therefore, a real, inverted and enlarged image of the object is formed beyond the centre of curvature.

Answer 2

Answer:

Focal length = 10cm

Object distance= 15cm

Height of the object = 6cm

Using the mirror's formula,

1/f = 1/v + 1/u

= 1/-10 = 1/v + 1/-15

= 1/v = 1/-10 + 1/15

= 1/v = (-3+2)/30

= v= -30

Since, image distance is negative,

Therefore, it is real and inverted.

Now, magnification = -v/u

=-(-30)/15

= 2

Since, magnification is greater than 1,

Therefore, image is magnified.

Also,

m= H of image/H of object

= 2 = H of image/6

Height of image =12 cm

Characteristics of image are =

Real, inverted and magnified.