The magnifying power of a compound microscope increases with:
A) The focal length of the objective lens is increased and that of the eye lens is decreased
B) The focal length of eye lens is increased and that of objective lens is decreased
C) Focal lengths of both objective and eye-piece are increased
D) Focal lengths of both objective and eye-piece are decreased
Answer
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Hint: This question is totally based on the understanding of the compound microscope and the images formed by it. We need to be very clear with the types of lenses used in compound microscopes and the images formed by it. Also, we need to know the criteria on which the magnifying power of the compound microscope depends.
Complete step by step solution:
Already we know that in a compound microscope, we use two lenses in which one of smaller focal length and the other with a larger focal lens. Both the lenses are converging in nature. The lens with smaller focal length is known as objective and the lens with larger focal length is known as eyepiece.
Now, we need to write the formula for the magnification of the compound microscope. So, we can write the required formula as,$m = - \dfrac{{LD}}{{{f_0}{f_e}}}$
Where, $L$ is the tube length of the compound microscope;
$D$ is the distance of least vision;
${f_o}$ is the focal length of the object
And ${f_e}$ is the focal length of the eyepiece.
As, we can see from step two, that both the focal length of eyepiece and objective are inversely proportional to the magnification of the compound lens. So, we can conclude that increasing the focal length of both the objective and eyepiece the value of magnification will be reduced and vice versa.
Hence, option (D), i.e. focal lengths of both objective and eye-piece are decreased, is the correct choice of the given question.
Note: We should know this fact that two lenses are used in a compound microscope and only one lens is used in a simple lens. The final image formed by the compound microscope is highly enlarged and virtual. The distance between the focal lengths of objective and eyepiece is known as tube length of the microscope.
Complete step by step solution:
Already we know that in a compound microscope, we use two lenses in which one of smaller focal length and the other with a larger focal lens. Both the lenses are converging in nature. The lens with smaller focal length is known as objective and the lens with larger focal length is known as eyepiece.
Now, we need to write the formula for the magnification of the compound microscope. So, we can write the required formula as,$m = - \dfrac{{LD}}{{{f_0}{f_e}}}$
Where, $L$ is the tube length of the compound microscope;
$D$ is the distance of least vision;
${f_o}$ is the focal length of the object
And ${f_e}$ is the focal length of the eyepiece.
As, we can see from step two, that both the focal length of eyepiece and objective are inversely proportional to the magnification of the compound lens. So, we can conclude that increasing the focal length of both the objective and eyepiece the value of magnification will be reduced and vice versa.
Hence, option (D), i.e. focal lengths of both objective and eye-piece are decreased, is the correct choice of the given question.
Note: We should know this fact that two lenses are used in a compound microscope and only one lens is used in a simple lens. The final image formed by the compound microscope is highly enlarged and virtual. The distance between the focal lengths of objective and eyepiece is known as tube length of the microscope.
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