Question

The objective and eyepiece of an astronomical telescope are doubled convex lenses with refractive index 1.5. When the telescope is adjusted to infinity, the separation between the 2 lenses is 16 cm. If the space between the lenses is now filled with water and again telescope is adjusted to infinity, then the present separation between the lenses is
(a) 8 cm
(b) 16 cm
(c) 24 cm
(d) 32 cm

Answer 1

Hint: At infinity adjustment, the image formed by the objective is at the focus of the eyepiece. Hence, use the lens maker formula to get focus with and without water insertion cases.

Formula used:
When focused at infinity, separation between lenses for telescope =sum of focal distance of eyepiece and objective =${f_e} + {f_o}$
Focal distance of normal human eye${f_e} = 2.5cm$(air)
Lens-Maker formula for interface between two different media:
$\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{({\mu _2} - {\mu _1})}}{{{R_1}}}$ …………….... (1)
Where,
${\mu _2}$ refractive index of 2nd medium (upcoming)
${\mu _1}$ refractive index of 1st medium (ongoing)
$v$ and $u$ are the final images and initial image position respectively.
${R_1}$ is the radius of curvature of 1st medium (ongoing)

Complete step by step answer:
Given:
1. Refractive index of glass ${\mu _g} = 1.5$
2. Separation between lenses $l = 16cm$
We know,
Refractive index of water, ${\mu _w} = 1.3$
Refractive index of water, ${\mu _a} = 1$
To find: Separation of lenses in the case when the space between the lenses is filled with water in normal adjustment.

Step 1 of 6:
Let’s find the focus ${f_0}$ of the objective piece by applying the lens maker formula on each side of the lens.
Final image formed objective lens by using superposition of image (using image formed by left as source for right side of lens)-
$\dfrac{{{\mu _w}}}{{{v^{}}}} - \dfrac{{{\mu _a}}}{u} = \dfrac{{({\mu _w} - {\mu _g})}}{{{R_2}}}$+$\dfrac{{({\mu _g} - {\mu _a})}}{{{R_1}}}$ …………..…… (2)

Step 2 of 6:
Applying equation (2) on objective, (air to air travel)
When no water inserted-
Putting, $v = {f_0}{\text{ and }}u = \infty $, by using sign convention -
Using equation (2) we get,
$\dfrac{{{\mu _a}}}{{{f_0}}} - \dfrac{{{\mu _a}}}{\infty } = \dfrac{{({\mu _a} - {\mu _g})}}{{ - {R_o}}}$+$\dfrac{{({\mu _g} - {\mu _a})}}{{{R_0}}}$
$ \Rightarrow \dfrac{1}{{{f_0}}} - \dfrac{1}{\infty } = \dfrac{{(1 - 1.5)}}{{ - {R_o}}}$+$\dfrac{{(1.5 - 1)}}{{{R_0}}}$
$ \Rightarrow \dfrac{1}{{{f_0}}} = \dfrac{{0.5 + 0.5}}{{{R_o}}} = \dfrac{1}{{{R_o}}}$
$ \Rightarrow {f_0} = {R_o}$

Step 3 of 6:
Calculating $R$ by using separation relation from the formula:
$ \Rightarrow {f_0} + {f_e} = 16cm \\
 \Rightarrow {f_0} = 16cm - 2.5cm \\
 \Rightarrow {f_0} = {R_o} = 13.5cm \\
$

Step 4 of 6:
Applying equation (2) on objective, (air to the glass to water travel)
When water inserted:
Putting, $v = {f_{0,new}}{\text{ and }}u = \infty ,{\text{ by using sign convention - }}$Using equation (2) we get
$ \Rightarrow \dfrac{{{\mu _w}}}{{{f_0}}} - \dfrac{{{\mu _a}}}{\infty } = \dfrac{{({\mu _w} - {\mu _g})}}{{ - {R_o}}}$+$\dfrac{{({\mu _g} - {\mu _a})}}{{{R_0}}}$
$\Rightarrow \dfrac{{1.33}}{{{f_{0,new}}}} - \dfrac{1}{\infty } = \dfrac{{(1.33 - 1.5)}}{{ - {R_o}}}$+$\dfrac{{(1.5 - 1)}}{{{R_0}}}$
$ \Rightarrow \dfrac{{1.33}}{{{f_{0,new}}}} = \dfrac{{0.17 + 0.5}}{{{R_o}}} = \dfrac{{0.67}}{{{R_o}}}$
$ \Rightarrow {f_{0,new}} = \dfrac{{1.33{R_o}}}{{0.67}} \approx 2{R_o} \approx 2{f_o}$

Step 5 of 6:
Realizing the fact that the system is symmetric about water medium- we can get
Using step 1: ${f_e} = {R_e}$
Using step 2: ${f_{e,new}} = 2{R_e} = 2{f_e}$

Step 6 of 6:
Hence, new separation is given by-
$
   = {f_{0,new}} + {f_{e,new}} = 2({f_o} + {f_e}) \\
   = 2 \times 16cm = 32cm \\
 $

Hence, option (d) 32cm is correct.

Note:
Here, few things should be understood clearly, which include, sign convention used for solving ray optics problems and apparent focal length changes when inserted in any medium. Do remember to take the initial distance to be infinity.