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 .

i think the answer would be b

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer