A compound microscope has an objective of focal length 1 cm and an eyepiece of focal length 2.5 cm

A compound microscope has an objective of focal length 1.25 cm and eyepiece of focal length 5 cm. A small object is kept at 2.5 cm from the objective. If the final image formed is at infinity, find the distance between the objective and the eyepiece ?

Distance between the objective and the eyepiece, L =

\[v_0 + \left| u_e \right|\]

To find v0, we have:

\[u_0 = - 2 . 5 \text {cm and } f_0 = 1 . 25 cm\]

\[\text { Now }, - \frac{1}{u_0} + \frac{1}{v_0} = \frac{1}{f_0}\] \[or \ v_0 = 2 . 5 cm\]

To find ue, we have:

\[v_e = \infty\text { and } f_e = 5 cm\]

Calculating using the same formula as above, we get:

\[u_e = - 5 cm\]

∴ L = 2.5 + 5 = 7.5 cm

Concept: Optical Instruments - The Microscope

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Text Solution

Solution : Here, `f_0 = 1 cm, f_e = 2.5 cm`. <br> `u_0 = -1.2 cm., m = ? L = ?` <br> As `(1)/(v_0)-(1)/(u_0)=(1)/(f_0)` <br> `:. (1)/(v_0)=(1)/(f_0)+(1)/(u_0)=(1)/(1)-(1)/(1.2)=(0.2)/(1.2)` <br> `v_0 = 1.2//0.2 = 6 cm` <br> As `m = (v_0)/(|u_0|)(1 + (d)/(f_e))` <br> `:. m = (6)/(1.2)(1 + (25)/(2.5)) = 55` <br> `L = v_0 + f_e = 6 + 2.5 = 8.5 cm`.