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

Vedclass · Accepted Answer

Increase the refractive index of the medium between the object and objective lens.

Vedclass · Answer

(B) The resolving power $(RP)$ of a compound microscope is given by the formula: $RP = \frac{2 \mu \sin 	heta}{1.22 \lambda}$,where $\mu$ is the refractive index of the medium between the object and the objective lens,$	heta$ is the half-angle of the cone of light from the object,and $\lambda$ is the wavelength of light used.To improve the resolving power,one must increase the numerator $(2 \mu \sin 	heta)$ or decrease the denominator $(1.22 \lambda)$.Increasing the refractive index $(\mu)$ of the medium (e.g.,using oil immersion) directly increases the resolving power. Therefore,option $B$ is the correct choice.

$A$ scientist is observing a bacteria through a compound microscope. For better analysis and to improve its resolving power,he should: (Select the best option)

Similar Questions

If the focal length of the objective and eye lens are $1.2 \, cm$ and $3 \, cm$ respectively,and the object is placed $1.25 \, cm$ away from the objective lens,and the final image is formed at infinity,the magnifying power of the microscope is:

In a compound microscope,the focal length and the aperture of the objective lens used are respectively:

What is a simple microscope? Obtain the equation of magnification for the image formed at the near point (normal vision distance).

In a simple microscope,if the final image is located at infinity,then its magnifying power is:

$A$ simple microscope is a combination of two lenses in contact. Their powers are $+20$ $D$ and $-4$ $D$. The distance of distinct vision is $25$ cm. When seen through the microscope, the size of an object $3$ mm high is (in $cm$)

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