Question

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

Answer 1

Given, height of object = 5cm

Position of object, u = - 25cm

Focal length of the lens, f = 10 cm

Hence, position of image, v =?

We know that,

1/v - 1/u = 1/f

1/v + 1/25 = 1/10

So, 1/v = 1/10 - 1/25

S0, 1/v = (5 - 2)/50

That is, 1/v = 3/50

So, v= 50/3 = 16.66 cm

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

Now, magnification = v/u

That is, m = 16.66/-25 = -0.66

Also, m= height of image/height of object

Or, -0.66 = height of image / 5 cm

Therefore, height of image = -3.3 cm

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

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

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

Nature of image – Real and inverted....

show how would you connect the three resistors, each of resistance 6 so that the combination has a resistance of 9 , 4