The power of a lens (biconvex) is $$1.25 \mathrm{~m}^{-1}$$ in particular medium. Refractive index of the lens is 1.5 and radii of curvature are $$20 \mathrm{~cm}$$ and $$40 \mathrm{~cm}$$ respectively. The refractive index of surrounding medium:

As shown in the figure, after passing through the medium 1 . The speed of light $$v_{2}$$ in medium 2 will be :

$$\left(\right.$$ Given $$\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$$ )

In normal adujstment, for a refracting telescope, the distance between objective and eye piece is $$30 \mathrm{~cm}$$. The focal length of the objective, when the angular magnification of the telescope is 2 , will be :

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 microscope's resolving power due to oil and choose the correct option.