Question

you can aswer both or any one these​

Answer 1

Explanation:

An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

Given that

The height of object = 5cm

Position of object, u = – 25cm

Focal length of the lens, f = 10 cm

We need to find

The position of image, v =?

Size of the image

Nature of the image

Formula

We know that

1/v – 1/u = 1/f

Substituting the known values in the above equation we get,

1/v + 1/25 = 1/10

=> 1/v = 1/10 – 1/25

=> 1/v = (5 – 2)/50

Hence, 1/v = 3/50

So, v= 50/3 = 16.66 cm

Therefore, the distance of the image is 16.66 cm on the opposite side of the lens.

Now, we know that

Magnification = v/u

Hence, m = 16.66/-25 = -0.66

Also, we know that

m= height of image/height of object

Or, -0.66 = height of image / 5 cm

Hence, height of image = -3.3 cm

The negative sign of height of the image depicts that an inverted image is formed.

So, the position of image = At 16.66 cm on the opposite side of the lens

Size of image = – 3.3 cm at the opposite side of the lens

Nature of image – Real and inverted

A concave lens of focal length 15cm forms an image 10 cm from the lens . How far is the object placed from the lens? Draw the ray diagram.

ANSWER

f = -15 cm

v = -10 cm

Lens formula: 1/v – 1/u =1/f

1/-10 – 1/u = 1/-15

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

1/u = -2/60

u = -30 cm

Object is at 30 cm from the concave lens (on left side).

Here, OB = 30 cm

OF = 15 cm

OB’ = 10 cm