# NEET Physics Ray Optics and Optical Instruments Questions Solved

BOARD

(i) What is the power of the lens whose focal length is 0.25m ?

(ii) The radii of curvature of the faces of a double convex lens are 20 cm and 25 cm. Its focal length is

20 cm. Calculate the refraction index of the material.                                 (2 marks)

(i)    Given focal lingth,

Power, $\mathrm{P}=\frac{1}{\mathrm{f}}$

(ii)    $\frac{1}{\mathrm{f}}=\left(\mathrm{u}-1\right)\left(\frac{1}{{\mathrm{R}}_{1}}-\frac{1}{{\mathrm{R}}_{2}}\right)$

$\mathrm{u}-1=\frac{1}{\mathrm{f}\left(\frac{1}{{\mathrm{R}}_{1}}-\frac{1}{{\mathrm{R}}_{2}}\right)}=\frac{1}{20\left(\frac{1}{20}+\frac{1}{25}\right)}$

$=\frac{5}{9}=0.55$

$\therefore \mathrm{u}=1.55$

Difficulty Level:

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