The length of the tube of a compound microscope is 15 cm. The focal length of objective and eye lenses are 1 cm and 5 cm respectively. What is the magnifying power of the microscope for the relaxed vision?
${\text{A}}{\text{.}}$ 50
${\text{B}}{\text{.}}$ 75
${\text{C}}{\text{.}}$ 25
${\text{D}}{\text{.}}$ 100
Answer
625.5k+ views
Hint: Here, we will proceed by using the formula for the calculation of the magnifying power of a compound microscope for relaxed vision in terms of length of the tube, focal lengths of the objective lens and eye lens.
Formula Used: ${\text{m}} = \dfrac{{{\text{L}} \times {\text{D}}}}{{{{\text{f}}_{\text{o}}} \times {{\text{f}}_{\text{e}}}}}$.
Complete Step-by-Step solution:
Given, Length of tube of compound microscope, L = 15 cm
Focal length of the objective lens, ${{\text{f}}_{\text{o}}}$ = 1 cm
Focal length of the eye lens, ${{\text{f}}_{\text{e}}}$ = 5 cm
As we know that the formula for the magnifying power of a compound microscope for relaxed vision is given by
Magnifying power of the microscope for relaxed vision, ${\text{m}} = \dfrac{{{\text{L}} \times {\text{D}}}}{{{{\text{f}}_{\text{o}}} \times {{\text{f}}_{\text{e}}}}}{\text{ }} \to {\text{(1)}}$ where L denotes the length of the tube of the compound microscope, ${{\text{f}}_{\text{o}}}$ denotes the focal length of the objective lens, ${{\text{f}}_{\text{e}}}$ denotes the focal length of the eye lens and D denotes the least distance for distinct vision which is equal to 25 cm for a normal eye (D = 25 cm)
By substituting L = 15 cm, ${{\text{f}}_{\text{o}}}$ = 1 cm, ${{\text{f}}_{\text{e}}}$ = 5 cm and D = 25 cm in the formula given by equation (1), we get
$ \Rightarrow {\text{m}} = \dfrac{{{\text{15}} \times {\text{25}}}}{{{\text{1}} \times {\text{5}}}} = 15 \times {\text{5}} = 75$
Therefore, the required magnifying power of the given compound microscope for the relaxed vision is 75.
Hence, option B is correct.
Note- A compound microscope is a microscope with high power (high magnification) which uses a compound lens system. To achieve a high magnification, a compound microscope has multiple lenses through which the objective lens is compounded by the eyepiece lens. Higher magnification is accomplished using two lenses instead of just one single magnifying lens.
Formula Used: ${\text{m}} = \dfrac{{{\text{L}} \times {\text{D}}}}{{{{\text{f}}_{\text{o}}} \times {{\text{f}}_{\text{e}}}}}$.
Complete Step-by-Step solution:
Given, Length of tube of compound microscope, L = 15 cm
Focal length of the objective lens, ${{\text{f}}_{\text{o}}}$ = 1 cm
Focal length of the eye lens, ${{\text{f}}_{\text{e}}}$ = 5 cm
As we know that the formula for the magnifying power of a compound microscope for relaxed vision is given by
Magnifying power of the microscope for relaxed vision, ${\text{m}} = \dfrac{{{\text{L}} \times {\text{D}}}}{{{{\text{f}}_{\text{o}}} \times {{\text{f}}_{\text{e}}}}}{\text{ }} \to {\text{(1)}}$ where L denotes the length of the tube of the compound microscope, ${{\text{f}}_{\text{o}}}$ denotes the focal length of the objective lens, ${{\text{f}}_{\text{e}}}$ denotes the focal length of the eye lens and D denotes the least distance for distinct vision which is equal to 25 cm for a normal eye (D = 25 cm)
By substituting L = 15 cm, ${{\text{f}}_{\text{o}}}$ = 1 cm, ${{\text{f}}_{\text{e}}}$ = 5 cm and D = 25 cm in the formula given by equation (1), we get
$ \Rightarrow {\text{m}} = \dfrac{{{\text{15}} \times {\text{25}}}}{{{\text{1}} \times {\text{5}}}} = 15 \times {\text{5}} = 75$
Therefore, the required magnifying power of the given compound microscope for the relaxed vision is 75.
Hence, option B is correct.
Note- A compound microscope is a microscope with high power (high magnification) which uses a compound lens system. To achieve a high magnification, a compound microscope has multiple lenses through which the objective lens is compounded by the eyepiece lens. Higher magnification is accomplished using two lenses instead of just one single magnifying lens.
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