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Question: Wavelength of light used in an optical instrument are \({\lambda _1} = 4000{A^ \circ }\) and \({\lam...

Wavelength of light used in an optical instrument are λ1=4000A{\lambda _1} = 4000{A^ \circ } and λ2=5000A{\lambda _2} = 5000{A^ \circ } then ratio of their respective resolving powers (corresponding to λ1{\lambda _1} and λ2{\lambda _2}​) is
a. 16:2516:25
b. 9:19:1
c. 4:54:5
d. 5:45:4

Explanation

Solution

Resolving power of a lens is known by its power differentiating between the objects. The larger the values of resolving power then smaller the value of distance between the lines. The resolving power is inversely proportional to the power. Apply this logic to get the ratios between the resolving powers.

Formula:
Resolving power α1λ\alpha \dfrac{1}{\lambda }, λ\lambda is the wavelength.

Complete step by step answer:
Resolving power of a lens is known by its power to differentiate between the objects. The larger the value of resolving powers then smaller the value of distance between the lines. The resolving power is inversely proportional to the power.

Bending of a light passing from one medium to another medium due to the change in speed is the Refraction. The ray bends towards the normal, when a light travels from a rarer medium to a denser medium. The magnification is the ratio of the height of an image to that of the height of the object. Lens formula gives the object distance, image distance and the focal length relationship.

The resolving power is inversely proportional to the power.
RPλ2RPλ1=λ1λ2\dfrac{{R{P_{\lambda 2}}}}{{R{P_{\lambda 1}}}} = \dfrac{{{\lambda _1}}}{{{\lambda _2}}}
Putting the values and we can write it as,
RPλ2RPλ1=40005000=45\dfrac{{R{P_{\lambda 2}}}}{{R{P_{\lambda 1}}}} = \dfrac{{4000}}{{5000}} = \dfrac{4}{5}
On simplification we get
RPλ1RPλ2=54\dfrac{{R{P_{\lambda 1}}}}{{R{P_{\lambda 2}}}} = \dfrac{5}{4}.

Hence, the correct answer is option (D).

Note: For better resolution the value of sinθ\sin \theta should be maximum. Material with higher refractive index must be used. The larger the values of resolving power then smaller the value of distance between the lines. This tells us that sometimes viruses, proteins can’t be imaged. The resolving power is inversely proportional to the power.