
A compound microscope has a magnifying power 48. The focal length of its eye-piece is 5cm. If final image is formed at least distance of distinct vision (25cm) then magnification of objective lens is
A. $-8$
B. $8$
C. $-200$
D. $200$
Answer
580.2k+ views
Hint:
A microscope is an optical instrument which forms a magnified image of very small objects held close to the eye. It has two convex lenses.
Formula used:
Magnification of compound microscope is given by,$m = {m_o} \times {m_e}$
Where, ${m_o}$ is magnification of an objective lens, ${m_e}$ is the magnification of the eye piece.
Magnification of the eyepiece is given by,${m_e} = 1 + \dfrac{D}{{{f_e}}}$
Where ${f_e}$is the focal length of the eye piece, D is the least distance if distinct vision.
Complete step by step answer:
Let us consider the magnification of the microscope.
Given that the magnification of compound microscope is m=-48
Magnification of objective lenses is negative. Since it produces real images.
Magnification of the eyepiece is positive. However, magnification of the compound microscope is negative ( though the image is virtual, it is inverted with respect to the object viewed).
Let us consider the focal length of the eye piece${f_e}$=5cm
We know that the Least distance if distinct vision=25cm
If the Magnification of the eyepiece is given by,${m_e} = 1 + \dfrac{D}{{{f_e}}}$
${m_e} = 1 + \dfrac{{25}}{5}$
$ = 6$
Then,
Magnification of compound microscope is given by ,$m = {m_o} \times {m_e}$
${m_o} = \dfrac{m}{{{m_e}}}$
$ = \dfrac{{ - 48}}{6}$
\[ = - 8\]
So, the correct answer is “Option A”.
Additional Information:
A microscope is an optical instrument that forms a magnified image of very small objects held close to the eye. There are mainly two types of microscopes.
Simple microscope
Compound microscope.
A compound microscope consists of two convex lenses of suitable focal lengths and forms highly magnified images of tiny objects. A compound microscope is based on the principle that a convex lens can form magnified images in the following two ways:
When the object is within the focal length of a convex lens, a magnified, virtual and erect image of the object is formed.
When the object is between the focal length f and 2f from the convex lens, the image formed is magnified, real, and inverted.
In a compound microscope, both these effects are used to produce a highly magnified image of tiny objects. It consists of two convex lenses fitted co-axially at the free ends of a long metallic tube. One convex lens is called the objective lens because placed near to the object, having a small focal length and small aperture. The other convex lens is called an eyepiece has a moderate focal length and aperture greater than that of the objective lens. The final image is viewed through the eyepiece.
Note:
For a normal eye, the near point is at a distance of 25cm. This distance is known as the least distance of distinct vision (D).
The closest point from the eye at which an object is seen clearly is called the near point of the eye.
For a normal eye, the far point is at infinity.
A microscope is an optical instrument which forms a magnified image of very small objects held close to the eye. It has two convex lenses.
Formula used:
Magnification of compound microscope is given by,$m = {m_o} \times {m_e}$
Where, ${m_o}$ is magnification of an objective lens, ${m_e}$ is the magnification of the eye piece.
Magnification of the eyepiece is given by,${m_e} = 1 + \dfrac{D}{{{f_e}}}$
Where ${f_e}$is the focal length of the eye piece, D is the least distance if distinct vision.
Complete step by step answer:
Let us consider the magnification of the microscope.
Given that the magnification of compound microscope is m=-48
Magnification of objective lenses is negative. Since it produces real images.
Magnification of the eyepiece is positive. However, magnification of the compound microscope is negative ( though the image is virtual, it is inverted with respect to the object viewed).
Let us consider the focal length of the eye piece${f_e}$=5cm
We know that the Least distance if distinct vision=25cm
If the Magnification of the eyepiece is given by,${m_e} = 1 + \dfrac{D}{{{f_e}}}$
${m_e} = 1 + \dfrac{{25}}{5}$
$ = 6$
Then,
Magnification of compound microscope is given by ,$m = {m_o} \times {m_e}$
${m_o} = \dfrac{m}{{{m_e}}}$
$ = \dfrac{{ - 48}}{6}$
\[ = - 8\]
So, the correct answer is “Option A”.
Additional Information:
A microscope is an optical instrument that forms a magnified image of very small objects held close to the eye. There are mainly two types of microscopes.
Simple microscope
Compound microscope.
A compound microscope consists of two convex lenses of suitable focal lengths and forms highly magnified images of tiny objects. A compound microscope is based on the principle that a convex lens can form magnified images in the following two ways:
When the object is within the focal length of a convex lens, a magnified, virtual and erect image of the object is formed.
When the object is between the focal length f and 2f from the convex lens, the image formed is magnified, real, and inverted.
In a compound microscope, both these effects are used to produce a highly magnified image of tiny objects. It consists of two convex lenses fitted co-axially at the free ends of a long metallic tube. One convex lens is called the objective lens because placed near to the object, having a small focal length and small aperture. The other convex lens is called an eyepiece has a moderate focal length and aperture greater than that of the objective lens. The final image is viewed through the eyepiece.
Note:
For a normal eye, the near point is at a distance of 25cm. This distance is known as the least distance of distinct vision (D).
The closest point from the eye at which an object is seen clearly is called the near point of the eye.
For a normal eye, the far point is at infinity.
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