
Least distance of distinct vision is 25cm. What will be the magnifying power of a simple microscope of focal length 5 cm, if the final image is formed at a minimum distance of distinct vision?
(A) 1/5
(B) 5
(C) 1/6
(D) 6
Answer
446.5k+ views
Hint: The magnifying power of a simple microscope, when the final image is at minimum distance of distinct vision can be calculated by $M = 1 + \dfrac{D}{f}$ , where D denotes the minimum distance of distinct vision and f is the focal length of the simple microscope. Since the image is formed at a minimum distance of distinct vision, we can use the formula in this case and focal length (f) is also given in the question. Using this information, we can calculate magnification.
Complete step by step solution:
The magnifying power of a simple microscope is given by $M = 1 + \dfrac{D}{f}$, when the final image is at minimum distance of distinct vision.
In the question, it is given that the least or minimum distance of distinct vision is 25 cm, and focal length of the microscope is 5 cm.
Putting these values in the above formula we get
$M = 1 + \dfrac{D}{f}$
$M = 1 + \dfrac{{25}}{5}$
$M = 1 + 5$
$M = 6$
Therefore, the correct answer is option D.
Additional information: The magnifying power of a simple microscope is given by $M = \dfrac{D}{u}$ , where u denotes the object distance from the microscope. Based on the distance of the image from the microscope, we can calculate the value $\dfrac{1}{u}$ using Lens formula \[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\] and substitute in the formula of magnification.
When the final image is at infinite distance from the microscope, i.e. $v = - \infty $, we get $\dfrac{1}{u} = \dfrac{1}{f}$ , so the magnification is $M = \dfrac{D}{f}$
When the final image is at minimum distance of distinct vision from the microscope, i.e. $v = - D$, we get $\dfrac{1}{u} = \dfrac{1}{f} + \dfrac{1}{D}$ , so the magnification is $M = 1 + \dfrac{D}{f}$
Note: Before solving the problem, the student needs to be able to understand the working of a simple microscope and remember the formulas used for the same. A common mistake that students make in such questions is using the formula for image at infinity in place of formula for image at minimum distance of distinct vision, or using the formula of minimum distance of distinct vision in place of formula for image at infinity.
Complete step by step solution:
The magnifying power of a simple microscope is given by $M = 1 + \dfrac{D}{f}$, when the final image is at minimum distance of distinct vision.
In the question, it is given that the least or minimum distance of distinct vision is 25 cm, and focal length of the microscope is 5 cm.
Putting these values in the above formula we get
$M = 1 + \dfrac{D}{f}$
$M = 1 + \dfrac{{25}}{5}$
$M = 1 + 5$
$M = 6$
Therefore, the correct answer is option D.
Additional information: The magnifying power of a simple microscope is given by $M = \dfrac{D}{u}$ , where u denotes the object distance from the microscope. Based on the distance of the image from the microscope, we can calculate the value $\dfrac{1}{u}$ using Lens formula \[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\] and substitute in the formula of magnification.
When the final image is at infinite distance from the microscope, i.e. $v = - \infty $, we get $\dfrac{1}{u} = \dfrac{1}{f}$ , so the magnification is $M = \dfrac{D}{f}$
When the final image is at minimum distance of distinct vision from the microscope, i.e. $v = - D$, we get $\dfrac{1}{u} = \dfrac{1}{f} + \dfrac{1}{D}$ , so the magnification is $M = 1 + \dfrac{D}{f}$
Note: Before solving the problem, the student needs to be able to understand the working of a simple microscope and remember the formulas used for the same. A common mistake that students make in such questions is using the formula for image at infinity in place of formula for image at minimum distance of distinct vision, or using the formula of minimum distance of distinct vision in place of formula for image at infinity.
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