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Draw a neat labelled diagram of a compound microscope and explain its working. Derive an expression for its magnification. |
Answer» Solution :Description : It consists of two convex lenses SEPARATED by a distance. The lens near the object is called objective and the lens near the eye is called eye piece. The objective lens has small focal length and eye piece has of larger focal length. The distance of the object can be ADJUSTED by means of a rack and pinion arrangement.  Working : The object OJ is placed outside the principal the principal focus of the objective and the real image is formed on the other side of it. The image `I_(1)G_(1)` is real, inverted and magnified. This image acts as the object for the eyepiece. The position of the eyepieceis so adjusted that the image due to the objectiveis between the optic centre and principal focus to form the final image at the near point. The final image IG is virtual, inverted and magnified. Magnifying Power : It is defined as the ratio of the angle subtended by the final image at the eye when formed at near point to the angle subtended by the object at the eye when imagined to be at near point. Imagining that the eye is at the optic centre, the angle subtended by the final image is `alpha`. When the object is imagined to be taken at near point it is represented by 1 J' and OJ = I J'. The angle made by I J' at the eye is `beta`. Then the definition of power `m=(alpha)/(beta)~=(tan alpha)/(tan beta)"for small angles"[{:(because Delta" IGO''" rArr Tan alpha=("IG")/("IO''")),(because "IJ'O''" rArrTan beta=("IJ'")/("IO''")):}]` `=("IG/IO''")/("IJ'/IO''")=("IG")/("IJ'")=("IG")/("OJ'")""(because "IJ' = OJ")` DIVIDING and multiplying by `I_(1)G_(1)` on the right side, we get `m=((IG)/(I_(1)G_(1)))((I_(1)G_(1))/(OJ))`. Magnifying power of the objective `(m_(0))=I_(1)G_(1)//OJ`= Height of the image due to the objective/Height of its object. Magnifying power of the eye piece `(m_(e))=IG//I_(1)G_(1)`=Height of the final image / Height of the object for the eyepiece. `THEREFORE m= m_(0)xxm_(e)"...............................(1)"` To find `m_(0)` : In figure OJ O' and `I_(1)G_(1)O'` are similar TRIANGLES. `((I_(1)G_(1))/(OG))=((O'I_(1))/(O'O))` Using sign convention, we find that `O'I_(1)=+v_(0) and O'O=-u` where `v_(0)` is the image distance due to the objective and u is the object distance for the objective or the compound microscope. `I_(1)G_(1)` is negative and OJ is positive. `therefore m_(0)=(v_(0))/(uu).(because (I_(1)G_(1))/(OJ)=m_(0))` To find `m_(e)` : The eyepiece behaves like a simple microscope. So the magnifying power of the eye piece. `therefore""m_(e)=(1+(D)/(f_(e)))` Where `f_(e)` is the focal length of the eyepiece. Substituting `m_(0)` and `m_(e)` in equation (1), `m=+(v_(0))/(u)(1+(D)/(f_(e)))` When the object is very close to the principal focus `F_(0)` of the objective, the image due to the objective becomes very close to the eyepiece. `u~~-f_(0) and v_(0) ~~L` Where L is the length of the microscope. Then `m~~-(L)/(f_(0))(1+(D)/(f_(e)))` |
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