Question

A compound microscope consists of an objective of focal length 1 cm and an eyepiece of focal length 5 cm. An object is placed at a distance of 0.5 cm from the objective. What should be the separation between the lenses so that the microscope projects an inverted real image of the object on a screen 30 cm behind the eyepiece?

Answer 1

Given :

Focal length of the objective, fo = 1.0 cm

Focal length of the eyepiece, fe = 5 cm

Distance of the object from the objective, u0 = 0.5 cm

Distance of the image from the eyepiece, ve = 30 cm

The lens formula for the objective lens is given by

1/v0-1/u0=1/f0

⇒1/v0+1/0.5=1/1

⇒ 1/v0=1-10/5=-1

⇒ v0=-1 cm

The objective will form a virtual image at the side same as that of the object at a distance of 1 cm from the objective lens. The image formed by the objective will act as a virtual object for the eyepiece.

The lens formula for the eyepiece is

1/ve-1/ue=1/fe

1/30-1/ue=1/5

⇒-1/ue=1/5-1/30

=6-1/30=1/6

⇒ ue = − 6 cm

∴ Separation between the objective and the eyepiece = 6 − 1 = 5 cm

Answer 2

ғᴏᴄᴀʟ ʟᴇɴɢᴛʜ ᴏғ ᴛʜᴇ ᴏʙᴊᴇᴄᴛɪᴠᴇ, ғᴏ = 1.0 ᴄᴍ

ғᴏᴄᴀʟ ʟᴇɴɢᴛʜ ᴏғ ᴛʜᴇ ᴇʏᴇᴘɪᴇᴄᴇ, ғᴇ = 5 ᴄᴍ

ᴅɪsᴛᴀɴᴄᴇ ᴏғ ᴛʜᴇ ᴏʙᴊᴇᴄᴛ ғʀᴏᴍ ᴛʜᴇ ᴏʙᴊᴇᴄᴛɪᴠᴇ, ᴜ0 = 0.5 ᴄᴍ

ᴅɪsᴛᴀɴᴄᴇ ᴏғ ᴛʜᴇ ɪᴍᴀɢᴇ ғʀᴏᴍ ᴛʜᴇ ᴇʏᴇᴘɪᴇᴄᴇ, ᴠᴇ = 30 ᴄᴍ

ᴛʜᴇ ʟᴇɴs ғᴏʀᴍᴜʟᴀ ғᴏʀ ᴛʜᴇ ᴏʙᴊᴇᴄᴛɪᴠᴇ ʟᴇɴs ɪs ɢɪᴠᴇɴ ʙʏ


1/ᴠ0-1/ᴜ0=1/ғ0

⇒1/ᴠ0+1/0.5=1/1


⇒ 1/ᴠ0=1-10/5=-1

⇒ ᴠ0=-1 ᴄᴍ

ᴛʜᴇ ᴏʙᴊᴇᴄᴛɪᴠᴇ ᴡɪʟʟ ғᴏʀᴍ ᴀ ᴠɪʀᴛᴜᴀʟ ɪᴍᴀɢᴇ ᴀᴛ ᴛʜᴇ sɪᴅᴇ sᴀᴍᴇ ᴀs ᴛʜᴀᴛ ᴏғ ᴛʜᴇ ᴏʙᴊᴇᴄᴛ ᴀᴛ ᴀ ᴅɪsᴛᴀɴᴄᴇ ᴏғ 1 ᴄᴍ ғʀᴏᴍ ᴛʜᴇ ᴏʙᴊᴇᴄᴛɪᴠᴇ ʟᴇɴs. ᴛʜᴇ ɪᴍᴀɢᴇ ғᴏʀᴍᴇᴅ ʙʏ ᴛʜᴇ ᴏʙᴊᴇᴄᴛɪᴠᴇ ᴡɪʟʟ ᴀᴄᴛ ᴀs ᴀ ᴠɪʀᴛᴜᴀʟ ᴏʙᴊᴇᴄᴛ ғᴏʀ ᴛʜᴇ ᴇʏᴇᴘɪᴇᴄᴇ.


ᴛʜᴇ ʟᴇɴs ғᴏʀᴍᴜʟᴀ ғᴏʀ ᴛʜᴇ ᴇʏᴇᴘɪᴇᴄᴇ ɪs

1/ᴠᴇ-1/ᴜᴇ=1/ғᴇ

1/30-1/ᴜᴇ=1/5

⇒-1/ᴜᴇ=1/5-1/30

=6-1/30=1/6

⇒ ᴜᴇ = − 6 ᴄᴍ

∴ sᴇᴘᴀʀᴀᴛɪᴏɴ ʙᴇᴛᴡᴇᴇɴ ᴛʜᴇ ᴏʙᴊᴇᴄᴛɪᴠᴇ ᴀɴᴅ ᴛʜᴇ ᴇʏᴇᴘɪᴇᴄᴇ = 6 − 1 = 5 ᴄᴍ