The lens shown in FIGURE CP$35.49$ is called an achromatic doublet, meaning that it has no chromatic aberration. The left side is flat, and all other surfaces have radii of curvature R.

a. For parallel light rays coming from the left, show that the effective focal length of this two-lens system is $f=R/12n_2-n_1-12$, where localid="1648757054673" $n_1$and $n_2$are, respectively, the indices of refraction of the diverging and the converging lenses. Don’t forget to make the thin-lens approximation.

b. Because of dispersion, either lens alone would focus red rays and blue rays at different points. Define $∆n_1$ and $∆n_2$ as $n_{blue}-n_{red}$ for the two lenses. What value of the ratio $∆n_1/∆n_2$makes $f_{blue}=f_{red}$for the two-lens system? That is, the two-lens system does not exhibit chromatic aberration.

c. Indices of refraction for two types of glass are given in the table. To make an achromatic doublet, which glass should you use for the converging lens and which for the diverging lens? Explain

$n_{blue}$ $n_{red}$

Crown glass 1.525 1.517

Flint glass 1.632 1.616

d. What value of R gives a focal length of $10.0cm$?

Step by step solution

01

Part (a) Step 1 : Given Information

We have to find out effective focal length of the two lens.

02

Part (a) Step 2 : Calculation

$$\begin{gathered} \frac{n_1}{s}+\frac{n_2}{s\text{'}}=\frac{n_2-n_1}{R} \\ \frac{n_1}{\infty }+\frac{n_2}{s\text{'}}=\frac{n_2-n_1}{R} \\ s\text{'}=R\frac{n_2}{n_2-n_1} \\ \frac{n_2}{R\frac{n_2}{n_2-n_1}}+\frac{n_{air}}{-f}=\frac{n_{air}-n_2}{R} \\ \frac{n_2-n_1}{R}+\frac{1}{-f}=\frac{1-n_2}{R} \\ \frac{1}{-f}=\frac{1-n_2-n_2+n_1}{R} \\ f=\frac{R}{2n_2-n_1-1} \end{gathered}$$

Here,$f$is focal length, $n$is refractive index and$R$is radius of curvature of lens

03

Part (b) Step 1 : Given Information

we have to find what value of the ratio$∆n_1/n_{2}makes{f}_{blue}=f_{red}$.

04

Part (b) Step 2 : Simplify

$$\begin{gathered} f_{blue}=\frac{R}{2{\left(n_2\right)}_{blue}-{\left(n_1\right)}_{blue}-1}and{f}_{red}=\frac{R}{2{\left(n_2\right)}_{red}-{\left(n_1\right)}_{red}-1} \\ Now,f_{blue}=f_{red}wegetratio: \\ {f}_{blue}=f_{red} \\ \frac{R}{2{\left(n_2\right)}_{blue}-{\left(n_1\right)}_{blue}-1}=\frac{R}{2{\left(n_2\right)}_{red}-{\left(n_1\right)}_{red}-1} \\ 2{\left(n_2\right)}_{blue}-{\left(n_1\right)}_{blue}-1=2{\left(n_2\right)}_{red}-{\left(n_1\right)}_{red}-1 \\ 2\left[{\left(n_2\right)}_{blue}-{\left(n_1\right)}_{red}\right]={\left(n_1\right)}_{blue}-{\left(n_1\right)}_{red}∆n_2=\frac{1}{2}∆n_1 \end{gathered}$$

Here,$f_{blue}$is focal length of blue and$n$is refractive index.

05

Part (c) Step 1 : Given Information

We have to find the$∆n$for each type of glass.

06

Part (c) Step 2 : Simplification 

$$\begin{gathered} ∆n_{crown}=1525-1517 \\ ∆n_{crown}=0.008 \\ ∆n_{flint}=1.632-1.616 \\ ∆n_{flint}=0.016 \end{gathered}$$.

Here,$n$is the refractive index for respective names.

07

Part (d) Step 1 : Given Information

We have to find focal length in expression for R.

08

Part (d) Step 2 : Explanation

$$\begin{gathered} n_1=1.632 \\ {n}_2=1.525 \\ f=10cm \\ So, \\ R=f(2n_2-n_1-1) \\ R=(10.0cm)\left[2(1.525)-1.632-1\right] \\ R=4.18cm \end{gathered}$$.

Here, $n$is refractive index and$f$is the focal length of lens.

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