Question

A ball is kept at a height h above the surface of a heavy transparent sphere made of a material of refractive index μ. The radius of the sphere is R. At t = 0, the ball is dropped to fall normally on the sphere. Find the speed of the image formed as a function of time fort<2hg. Consider only the image by a single refraction.

Answer 1

Given :

A transparent sphere refractive index= μ

radius of sphere =R

t=0 sec[ free fall-initially]

u= 0 m/s

let t be the time taken to travel the distance A to B.

Distance covered during this time is :

h=1/2 gt²

we are assuming distance of object from lens at any time is t.

u = -(h-1/2 gt²)

Refractive index of air, μ1 =  1

Refractive index of sphere, μ2 =  μ (given)

μ/V+ 1/((h-1/2gt²)= μ-1/R

μ/v=μ-1/R-1/h-1/2gt²=[(μ-1)h-1/2gt²-R] / R(h-12gt²)

Let v be the image distance at any time t. Then,

v=μR(h-1/2gt²)/ (μ-1)(h-1/2gt²-R)

Therefore, velocity of the image   V is given by,

V=dv/dt=d/dt [ μR(h-1/2gt²) / [(μ-1)h-1/2gt²]-R

=[ μR²gt / (μ-1)h-1/2gt²-R²]

Answer 2

ʀᴇғᴇʀ ᴛᴏ ᴛʜᴇ ᴀᴛᴛᴀᴄʜᴍᴇɴᴛ-------

ᴀ ᴛʀᴀɴsᴘᴀʀᴇɴᴛ sᴘʜᴇʀᴇ ʀᴇғʀᴀᴄᴛɪᴠᴇ ɪɴᴅᴇx= μ

ʀᴀᴅɪᴜs ᴏғ sᴘʜᴇʀᴇ =ʀ

ᴛ=0 sᴇᴄ[ ғʀᴇᴇ ғᴀʟʟ-ɪɴɪᴛɪᴀʟʟʏ]

ᴜ= 0 ᴍ/s

ʟᴇᴛ ᴛ ʙᴇ ᴛʜᴇ ᴛɪᴍᴇ ᴛᴀᴋᴇɴ ᴛᴏ ᴛʀᴀᴠᴇʟ ᴛʜᴇ ᴅɪsᴛᴀɴᴄᴇ ᴀ ᴛᴏ ʙ.

ᴅɪsᴛᴀɴᴄᴇ ᴄᴏᴠᴇʀᴇᴅ ᴅᴜʀɪɴɢ ᴛʜɪs ᴛɪᴍᴇ ɪs :

ʜ=1/2 ɢᴛ²

ᴡᴇ ᴀʀᴇ ᴀssᴜᴍɪɴɢ ᴅɪsᴛᴀɴᴄᴇ ᴏғ ᴏʙᴊᴇᴄᴛ ғʀᴏᴍ ʟᴇɴs ᴀᴛ ᴀɴʏ ᴛɪᴍᴇ ɪs ᴛ.

ᴜ = -(ʜ-1/2 ɢᴛ²)

ʀᴇғʀᴀᴄᴛɪᴠᴇ ɪɴᴅᴇx ᴏғ ᴀɪʀ, μ1 =  1

ʀᴇғʀᴀᴄᴛɪᴠᴇ ɪɴᴅᴇx ᴏғ sᴘʜᴇʀᴇ, μ2 =  μ (ɢɪᴠᴇɴ)

μ/ᴠ+ 1/((ʜ-1/2ɢᴛ²)= μ-1/ʀ

μ/ᴠ=μ-1/ʀ-1/ʜ-1/2ɢᴛ²=[(μ-1)ʜ-1/2ɢᴛ²-ʀ] / ʀ(ʜ-12ɢᴛ²)

ʟᴇᴛ ᴠ ʙᴇ ᴛʜᴇ ɪᴍᴀɢᴇ ᴅɪsᴛᴀɴᴄᴇ ᴀᴛ ᴀɴʏ ᴛɪᴍᴇ ᴛ. ᴛʜᴇɴ,

ᴠ=μʀ(ʜ-1/2ɢᴛ²)/ (μ-1)(ʜ-1/2ɢᴛ²-ʀ)

ᴛʜᴇʀᴇғᴏʀᴇ, ᴠᴇʟᴏᴄɪᴛʏ ᴏғ ᴛʜᴇ ɪᴍᴀɢᴇ   ᴠ ɪs ɢɪᴠᴇɴ ʙʏ,

ᴠ=ᴅᴠ/ᴅᴛ=ᴅ/ᴅᴛ [ μʀ(ʜ-1/2ɢᴛ²) / [(μ-1)ʜ-1/2ɢᴛ²]-ʀ

=[ μʀ²ɢᴛ / (μ-1)ʜ-1/2ɢᴛ²-ʀ²]