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The resolving power of a microscope is |
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Answer» Solution :The diagram related to the calculation of resolution of microscope. A microscope is used to see the details of the OBJECT under OBSERVATION. The ability of microscope depends not only in magnifying the object but also in resolving two points on the object separated by a small distance `d_(min)` Smaller the value of ` d_(min)` better will be the resolving power of the microscope. The radius of central MAXIMA us, `r_(0) = (1.22 lambdav)/(a)` In the place of focal lenghtf we have the image distance v. If the difference between the two points on the object to be resolved is `d_(min)` , then the magnicifation m is, `m = (r_(0))/(d_(min))` `d_(min)=(r_(0))/(am)=(1.22lambdau)/(a) [thereforem=v//u]` `d_(min)=(1.22lambdaf)/(a)[thereforeuapproxf]` On the object side, `2 tan beta approx = 2 sin beta = (a)/(f) "" THEREFORE [a = f 2sin beta]` `d_(min) = (1.22lambda)/(2sin beta)` The further reduce the value of `d_(min)` the OPTICAL pathof the light is increased by immersing the objective of the microscope in the a bath containing oil of refractive index n. 
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