Question

An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm find the position and nature of the image​

Answer 1

Question

An object is placed at a distance of 10cm from a convex mirror of focal length 15cm. Find the position and nature of the image ?

ANSWER

Given :-

An object is placed at a distance of 10cm from a convex mirror of focal length 15cm

Required to find :-

Position and nature of the image ?

Solution :-

Object distance (u) = -10cm

This is due to the rule of sign convention ! as all distance are measured with respective the pole.

Focal of a convex mirror is always positive !

So,

Focal length (f) = 15cm

Now,

We know that;

(1)/(f) = (1)/(u)+(1)/(v)

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

(1)/(v) = (1)/(15) - (1)/(-10)

(1)/(v) = (1)/(15)+(1)/(10)

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

(1)/(v) = (5)/(30)

v = (30)/(5)

v = 6 cm

Hence,

Image distance (v) = 6 cm

By drawing the ray diagram for the above situation using the information formulated from the above calculations. We can say that;

The position of the image between Pole and Focus.

The nature of image is Erect because it forms an upright image, Virtual because image is formed by diverging of the light rays, Image is considered to be as diminished as the size of the image is relatively smaller than the height of the object.

Answer 2

Answer:

Given that

u= -10 cm

f= 15 cm

We need to find v

We know the formula

1/f = 1/v + 1/u

Therefore, 1/v= 1/f – 1/u

Substituting the known values in the above equation we get,

1/v= 1/15 + 1/10

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

Hencev=5/30 = 6 cm

v= 6cm

Nature of the image: The image is formed behind the mirror and it is virtual and erect

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