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 :

Two coherent sources of light interfere. The intensity ratio of two sources is $$1: 4$$. For this interference pattern if the value of $$\frac{I_{\max }+I_{\min }}{I_{\max }-I_{\min }}$$ is equal to $$\frac{2 \alpha+1}{\beta+3}$$, then $$\frac{\alpha}{\beta}$$ 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.