
A combination is made of two lenses of focal lengths $f$ and ${f}'$ in contact; the dispersive powers of the materials of the lenses are $\omega$ and ${\omega }'$. The combination is achromatic when
$\begin{align}
& A.\omega ={{\omega }_{0}},{\omega }'=2{{\omega }_{0}},{f}'=2f \\
& B.\omega ={{\omega }_{0}},{\omega }'=2{{\omega }_{0}},{f}'=\dfrac{f}{2} \\
& C.\omega ={{\omega }_{0}},{\omega }'=2{{\omega }_{0}},{f}'=\dfrac{-f}{2} \\
& D.\omega ={{\omega }_{0}},{\omega }'=2{{\omega }_{0}},{f}'=-2f \\
\end{align}$
Answer
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Hint: An achromatic lens can be described as a lens which is created by the combination of two different kinds of lenses which are having different focal powers in a manner so that the images created by the light of both the combined lenses are free from chromatic aberration or otherwise known as achromatism. The equation for the condition of achromatism is,
$\dfrac{{{\omega }_{1}}}{{{f}_{1}}}+\dfrac{{{\omega }_{2}}}{{{f}_{2}}}=0$
Using this equation, we can find out the relation between the dispersive powers of both the lenses.
Complete answer:
For the achromatic combination of lenses, the condition of achromatism is given by the equation,
$\dfrac{{{\omega }_{1}}}{{{f}_{1}}}+\dfrac{{{\omega }_{2}}}{{{f}_{2}}}=0$
As per the question, it is already mentioned that the dispersive powers of the lenses are $\omega$ and ${\omega }'$ and the focal length of the lenses are given as $f$ and ${f}'$ respectively.
Substituting these in this equation will give,
$\dfrac{\omega }{f}+\dfrac{{{\omega }'}}{{{f}'}}=0$
Substitute the conditions given in each option in the equation,
\[\begin{align}
& \dfrac{{{\omega }_{0}}}{f}+\dfrac{2{{\omega }_{0}}}{{{f}'}}=0 \\
& \therefore {f}'=-2f \\
\end{align}\]
So, the correct answer is “Option D”.
Note:
An achromatic lens is basically a combination of convex and concave pieces of glass that focuses the different colour with different wavelengths in light into a single plane. Each kind of glass disperses each colour differently when put together, they will counterbalance each other and create a sharp image. Chromatic aberration is referred to the blurring or colouring of red, green, blue, yellow, purple, orange and so on, near the edges of an image. These colours usually seemed to appear in high-contrast photos with a lot of highlights and shadows.
$\dfrac{{{\omega }_{1}}}{{{f}_{1}}}+\dfrac{{{\omega }_{2}}}{{{f}_{2}}}=0$
Using this equation, we can find out the relation between the dispersive powers of both the lenses.
Complete answer:
For the achromatic combination of lenses, the condition of achromatism is given by the equation,
$\dfrac{{{\omega }_{1}}}{{{f}_{1}}}+\dfrac{{{\omega }_{2}}}{{{f}_{2}}}=0$
As per the question, it is already mentioned that the dispersive powers of the lenses are $\omega$ and ${\omega }'$ and the focal length of the lenses are given as $f$ and ${f}'$ respectively.
Substituting these in this equation will give,
$\dfrac{\omega }{f}+\dfrac{{{\omega }'}}{{{f}'}}=0$
Substitute the conditions given in each option in the equation,
\[\begin{align}
& \dfrac{{{\omega }_{0}}}{f}+\dfrac{2{{\omega }_{0}}}{{{f}'}}=0 \\
& \therefore {f}'=-2f \\
\end{align}\]
So, the correct answer is “Option D”.
Note:
An achromatic lens is basically a combination of convex and concave pieces of glass that focuses the different colour with different wavelengths in light into a single plane. Each kind of glass disperses each colour differently when put together, they will counterbalance each other and create a sharp image. Chromatic aberration is referred to the blurring or colouring of red, green, blue, yellow, purple, orange and so on, near the edges of an image. These colours usually seemed to appear in high-contrast photos with a lot of highlights and shadows.
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