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Light of wavelength 589 nm is used to view an object under a microscope. The aperature of the objective has a diameter of 0.900 cm. Find What effect would this have on the resolving power. If water (mu = 1.33) fills the space between the object and objective. |
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Answer» Solution :Wavelength of light in water, `lambda_(w)=(lambda_a)/(mu)= (589nm)/(1.33)= 443nm` `triangle theta = 1.22((443xx10^(-9)m)/(0.900xx10^(-2)m)) RAD` `= 6.00xx 10^(-5)rad` SINCE `triangle theta` in this case is LESS than that in case (a), resolving power increases. |
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