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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 (a) The limiting angle of resolution. (b) What is the maximum limit of revolution for this microscope Using visible light of any wavelength you desire (c) 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 :Here `lamda = 589 nm = 589 xx 10^(-9) m` `a = 0.900 cm = 0.900 xx 10^(-2) m` `Delta theta = 1.22 (lamda/a) = 1.22 ((589 xx 10^(-9)m)/(0.900 xx 10^(-2)m) ) rad` ` = 7.89 xx 10^(-5) rad ` (B)To obtain the smallest angle corresponding to the maximum limit of resolution, we have to use the shortest wavelength (`lamda` = 400nm) in the visible spectrum. Limiting angle of resolution `Delta theta = 1.22 ((400 xx 10^(-9) m)/(0.900 xx 10^(-2) m) )= rad = 5.42 xx 10^(-5) rad ` (c ) Wavelength of light in water, `lamda_w= (lamda_a)/(MU) = (589 nm)/(1.33) = 443nm ` `theta = 1.22 ((443 xx 10^(-9)m)/(0.900 xx 10^(-2) m) ) rad = 6.00 xx 10^(-5) rad ` Since `Delta theta `in this case is less than that in case (a), resolving power increases. |
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