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A microscope has objective of aperture 8mm and focal length 2.5 cm. Estimate its resolving power. Given lambda= 5500 dotA. |
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Answer» Solution :We assume that the object is placed a little beyond its focal distance, say `2.5 CM`. `tan ALPHA = (4 mm)/(2.5 cm)= (0.4)/(2.5)=0.16` Since `sin alpha` is small, `sin alpha approx tan alpha = 0.16` Thus, `triangle x =(1.22 lambda)/(2 mu sin alpha)=(1.22(5500xx 10^(-10)m))/(2xx 1xx 0.16) approx 2xx 10^(-6)m`.  .
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