Define the magnifying power of a compound microscope when the final image is formed at infinity

Define the magnifying power of a compound microscope when the final image is formed at infinity. Why must both the objective and the eyepiece of a compound microscope has short focal lengths? Explain.

When the final image is formed at infinity.
When the final image is formed at infinity, the angular magnification due to the eyepiece is

me = D/fe

Thus, the total magnification when the image is formed at infinity can be defined as the product of magnification of objective lens and eyepiece. i.e

m = mome = (L/fo) (D/fe)

From the above the equation, we can see that to achieve a large magnification of a small object, the objective and eyepiece should have small focal lengths.

Concept: Optical Instruments - The Microscope

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A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of a focal length 20 cm, and (b) a concave lens of focal length 16 cm?

Here, the point P on the right of the lens acts as a virtual object.

Object distance, u = 12 cm

∴           1v =120+112      =3+560 =860

i.e.,           v = 60/8 = 7.5 cm. 

Image is at a distance of 7.5 cm to the right of the lens, where the beam converges. (b)Now,Focal length of concave lens, f = –16 cmObject distance, u = 12 cm 

⇒      v = 48 cm 

Hence, the image is at a distance of 48 cm to the right of the lens, where the beam would converge.

Text Solution

Solution : Magnifying Power : The magnifying power of a compound microscope is denned as the ratio of the angle subtended at the eye by the finalimage to the angle subtended at the eye by the object, when both the final image and the object are situated at the least distance of distinct vision from the eye. <br> Magnifying power of the compound microscope when the final image is at infinity. <br> `m = -L/(f_(0)) xx D/(f_(e))` <br> From the above equation, we can see that to achieve a large magnification, the objective and eyepiece should have small focal lengths.

Text Solution

Solution : The angular magnification (or magnifying power) of a microscope is defined as the ratio of the angle subtended by the final image at the eye to the angle subtended by the object at the eye when seen directly. Angular magnification of a compound microscope when final image in a microscope is formed at infinity, is given by the relation :<br> `m=-L/f_(0).D/f_(e)` <br> Here L = length of microscope, D = least distance of distinct vision, `f_0` = focal length of objective lens and `f_(e)` = focal length of eye piece. <br> As magnifying power m is inversely proportional to both `f_0` and `f_(e)`, hence we prefer both objective and eyepiece to be of short focal lengths so as to have a high value of magnifying power

Text Solution

Solution : The magnification of compound microscope when the final image is formed at infinity, `m=L/(f_o)((D)/(f_e))` <br> For producing large magnifying power, both the objective and the eyepiece of a compound microscope should possess short focal length. <br> when ` f_(o) lt lt , f_e lt lt D,` <br> `m=(L/(f_o)-1)(1+D/(f_e)) ~~(L/(f_o)*D/(f_e))`