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Optics — Physics STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 SciencePhysicsOptics4 Marks
Question
(i) Explain chromatic aberration for spherical lenses. State a method to minimize or eliminate it.
(ii) What is achromatism? Derive a condition to achieve achromatism for a lens combination. State the conditions for it to be converging.
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Answer

(i)Lenses are prepared by using a transparent material medium having different refractive index for different colours. Hence angular dispersion is present.
If the lens is thick, this will result into notably different foci corresponding to each colour for a polychromatic beam, like a white light. This defect is called chromatic aberration.
As violet light has maximum deviation, it is focussed closest to the pole.

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Reducing/eliminating chromatic aberration:
  1. Eliminating chromatic aberrations for all colours is impossible. Hence, it is minimised by eliminating aberrations for extreme colours.
  2. This is achieved by using either a convex and a concave lens in contact or two thin convex lenses with proper separation. Such a combination is called achromatic combination.
    (ii)
    i. To eliminate chromatic aberrations for extreme colours from a lens, either a convex and a concave lens in contact or two thin convex lenses with proper separation are used.
    ii. This combination is called achromatic combination. The process of using this combination is termed as achromatism of a lens.
    iii. Let $ω_1$​​​​​​​ and $ω_2$​​​​​​​ be the dispersive powers of materials of the two component lenses used in contact for an achromatic combination.
    iv. Let $V, R$ and $Y$ denote the focal lengths for violet, red and yellow colours respectively.
    v. For lens $1$, let
    $K _1=\left(\frac{1}{R_1}-\frac{1}{R_2}\right)_1$ and $K _2=\left(\frac{1}{R_1}-\frac{1}{R_2}\right)_2$
    vi. For the combination to be achromatic, the resultant focal length of the combination must be the same for both the colours,
    $\therefore \quad f_V=f_R$
    $\therefore \quad \frac{1}{f_v}=\frac{1}{f_R}$
    For two thin lenses in contact, $\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$
    $\therefore \quad \frac{1}{\left( f _{ v }\right)_1}+\frac{1}{\left( f _{ v }\right)_2}=\frac{1}{\left( f _{ R }\right)_1}+\frac{1}{\left( f _{ R }\right)_2}$
    vii. From Lens maker's formula,
    $ {\left[\left( n _{ V }\right)_1-1\right] K _1+\left[\left( n _{ v }\right)_2-1\right] K _2}$
    $ =\left[\left( n _{ R }\right)_1-1\right] K _1+\left[\left( n _{ R }\right)_2-1\right] K _2$
    $\therefore \quad  \frac{ K _1}{ K _2}=\frac{\left( n _{ v }\right)_2-\left( n _{ R }\right)_2}{\left( n _{ v }\right)_1-\left( n _{ R }\right)_1}$
    $\text { viii. }  \text { Similarly, for mean colour (yellow), }$
    $ \frac{1}{ f _{ r }}=\frac{1}{\left( f _{ r }\right)_1}+\frac{1}{\left( f _{ r }\right)_2}$
    $\frac{ K _1}{ K _2}=\frac{\left( n _{ Y }\right)_2-1}{\left( n _{ Y }\right)_1-1} \times \frac{\left( f _{ v }\right)_2}{\left( f _{ Y }\right)_1}$
    ix. From equations (1) and (3),
    $\frac{\left(f_{ r }\right)_2}{\left(f_{ Y }\right)_1}=-\frac{\left( n _{ V }\right)_2-\left( n _{ R }\right)_2}{\left( n _{ V }\right)_1-\left( n _{ R }\right)_1} \times \frac{\left( n _{ Y }\right)_2-1}{\left( n _{ Y }\right)_1-1}$
    Now, dispersive power $\omega_1=\frac{\left( n _{ V }\right)_2-\left( n _{ k }\right)_2}{\left( n _{ Y }\right)_2-1}$ and
    $\omega_2=\frac{\left( n _{ v }\right)_2-\left( n _{ n }\right)_2}{\left( n _{ Y }\right)_2-1}$
    x. Substituting values of $\omega_1$ and $\omega_2$ in equation (4), we get,
    $\frac{\left(f_{ Y }\right)_2}{\left(f_{ Y }\right)_1}=-\frac{\omega_2}{\omega_1}$
    This is the condition for achromatism of a combination of lenses.
    Condition for converging:
    For this combination to be converging, $fY$ must be positive.
    Using equation $(3)$, for $f_Y$​​​​​​​ to be positive, $(f_Y)_1 < (f_Y)_2 \Rightarrow ω_1 < ω_2$​​​​​​​

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