Question

if an object of high 6cm is placed infront of a concave mirror at a distance of 20 cm with focal length 15 cm then find the image distance and characterstits​

Answer 1

Given:-

Object Distance (u) = - 20 cm

Focal length (f) = -15 cm

Height of object (h) = 6 cm

To find:-

Image distance (v)

Characteristics of the image.

Solution:-

Using Mirror Formula:-

$\boxed{\rm{ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} }}$

$\rm \frac{1}{ - 15} = \frac{1}{v} - \frac{1}{( - 20)}$

$\rm\frac{1}{v} = \frac{1}{ - 15} - \frac{1}{20}$

$\rm \frac{1}{v} = \frac{ - 4 - 3}{60}$

$\rm\frac{1}{v} = \frac{ - 7}{60}$

$\rm \red {v = \frac{ - 60}{7} =} \red{ - 8.57 \: cm}$

Characteristics of the image = Real and image is formed in front of mirror.

Answer 2

Given :-

◉ Object distance, u = -20cm

Object distance is always taken negative.

◉ Height of image, h₁ = +6cm

◉ Focal length, f = -15cm

Focal length of a concave mirror is always taken negative.

To Find :-

◉ Image distance and properties of image.

Solution :-

Using the mirror formula, we have:

⇒ 1/f = 1/v + 1/u

⇒ -1/15 = 1/v - 1/20

⇒ 1/v = 1/20 - 1/15

⇒ 1/v = 3 - 4/60

⇒ 1/v = -1/60

⇒ v = -60cm

Negative image distance means that the image is on the side of object so It must be real because In concave mirror Virtual image is formed behind the mirror with positive distance.

Now, we know

⇒ Magnification = -v/u = h₂ / h₁

Here, h₂ = height of image

⇒ -(-60/-30) = h₂ / 6

⇒ -2 = h₂ / 6

⇒ h₂ = -12cm

It is now clear that the image is:

• Double the size of object.
• Real and inverted.

Some Information :-

☛ Measures against the direction of incident rays are taken negative while measurements in the direction of incident rays are taken positive.

☛ Distance measured above the principle axis are taken positive while those measured below the principle axis are taken negative.