
An object is seen first in red light and then in violet light through a simple microscope. In which case is the magnification larger?
Answer
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Hint:Magnification of magnifying power is inversely proportional to focal distance. Focal length is inversely proportional to refractive index. Refractive index is inversely proportional to wavelength. Use all this relation and deduce a relation between magnification and wavelength.
Complete step by step answer:Using the above hint, we need to find a relation between magnification and wavelength. We know that the wavelength of red light is greater than the wavelength of violet light. If we can achieve a relation between magnification and wavelength then we can easily answer this question.
We know that magnification is given as:
$M = 1 + \dfrac{d}{f}$
Where $M$ is the magnification
$d$ is the distance between the lens and object
$f$ is the focal distance
Since the object which is seen in red light first and then in violet light will have the same distance from the magnifying lens. Hence the magnification is inversely proportional to focal distance.
$M\alpha \dfrac{1}{f}$
Focal length is inversely proportional to the refractive index of the medium, this is obtained from the makers formula. From this we have:
$M\alpha \dfrac{1}{f}\alpha \mu $
Where $\mu $ is the refractive index of the medium
Now, we also have that refractive index is inversely proportional to wavelength:
$ \Rightarrow M\alpha \dfrac{1}{f}\,\alpha \,\mu \,\alpha \dfrac{1}{\lambda }$
$ \Rightarrow M\,\alpha \,\dfrac{1}{\lambda }$
It is now clear that the magnification will be larger for light with smaller wavelengths.
The wavelength of violet light is less than the wavelength of red light.
Therefore, the magnification will be larger than the violet light.
Note:This is a difficult question. Remember the steps, how the magnification is related to focal distance and focal distance is related to refractive index. Finally, the refractive index is related to wavelength. Remember that the wavelength of violet light is smaller than the wavelength of red light.
Complete step by step answer:Using the above hint, we need to find a relation between magnification and wavelength. We know that the wavelength of red light is greater than the wavelength of violet light. If we can achieve a relation between magnification and wavelength then we can easily answer this question.
We know that magnification is given as:
$M = 1 + \dfrac{d}{f}$
Where $M$ is the magnification
$d$ is the distance between the lens and object
$f$ is the focal distance
Since the object which is seen in red light first and then in violet light will have the same distance from the magnifying lens. Hence the magnification is inversely proportional to focal distance.
$M\alpha \dfrac{1}{f}$
Focal length is inversely proportional to the refractive index of the medium, this is obtained from the makers formula. From this we have:
$M\alpha \dfrac{1}{f}\alpha \mu $
Where $\mu $ is the refractive index of the medium
Now, we also have that refractive index is inversely proportional to wavelength:
$ \Rightarrow M\alpha \dfrac{1}{f}\,\alpha \,\mu \,\alpha \dfrac{1}{\lambda }$
$ \Rightarrow M\,\alpha \,\dfrac{1}{\lambda }$
It is now clear that the magnification will be larger for light with smaller wavelengths.
The wavelength of violet light is less than the wavelength of red light.
Therefore, the magnification will be larger than the violet light.
Note:This is a difficult question. Remember the steps, how the magnification is related to focal distance and focal distance is related to refractive index. Finally, the refractive index is related to wavelength. Remember that the wavelength of violet light is smaller than the wavelength of red light.
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