A student focussed the image of a candle flame on a white screen using a convex lens. He noted down the position of candle, screen and the lens as under Position of candle = 12.0 cm Position of convex lens = 50.0 cm Position of the screen = 88.0 cm i. What is the focal length of the convex lens? ii. Where will the image be formed if he shifts the candle towards the lens at a position of 31.0 cm? iii. What will be the nature of the image formed if he further shifts the candle towards the lens? iv. Draw a ray diagram to show the formation of the image in case (iii) as said above.

Explanation:

Object distance = distance between object and lens

u = 50 - 12 = 38 cm

image distance v = 88 - 50 = 38 cm

i )

using lens formula

1 / v - 1 / u = 1 / f

1 / 38 - 1 / - 38 = 1/f

2 / 38 = 1  / f

f = 19 cm

ii ) So object is placed at 2 f or at the center of curvature of the lens .

If object distance is changed to 50 - 31 = 19 cm

using lens formula

1 / v - 1 / u = 1 / f

1 / v  + 1 /  19 = 1/19

1  / v = 0

v = infinity

It is so because now the object is placed at the focal distance

iii )

If object  further shifts towards the lens , then virtual image of object will be formed behind the lens which will be larger than object .

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.