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(B) The resolving power $(R.P.)$ of a microscope is given by the formula: $R.P. = \frac{2 \mu \sin 	heta}{1.22 \lambda}$,where $\mu$ is the refractiv

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Resolving power will be twice in the oil than it was in the air.

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(B) The resolving power $(R.P.)$ of a microscope is given by the formula: $R.P. = \frac{2 \mu \sin 	heta}{1.22 \lambda}$,where $\mu$ is the refractive index of the medium,$	heta$ is the semi-vertical angle,and $\lambda$ is the wavelength of light in vacuum/air.Initially,in air $(\mu_1 = 1)$: $(R.P.)_{	ext{air}} = \frac{2 	imes 1 	imes \sin 	heta}{1.22 \lambda} = \frac{2 \sin 	heta}{1.22 \lambda}$.When immersed in oil $(\mu_2 = 2)$: The wavelength in the medium becomes $\lambda_{	ext{oil}} = \frac{\lambda}{\mu_2} = \frac{\lambda}{2}$.Substituting this into the formula: $(R.P.)_{	ext{oil}} = \frac{2 \sin 	heta}{1.22 \lambda_{	ext{oil}}} = \frac{2 \sin 	heta}{1.22 (\lambda / 2)} = \frac{2 	imes 2 \sin 	heta}{1.22 \lambda} = 2 	imes (R.P.)_{	ext{air}}$.Therefore,the resolving power in oil is twice the resolving power in air.

$A$ microscope was initially placed in air (refractive index $1$). It is then immersed in oil (refractive index $2$). For a light whose wavelength in air is $\lambda$,calculate the change of the microscope's resolving power due to oil and choose the correct option.

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