An object is placed

Question:

An object is placed beyond the centre of curvature $\mathrm{C}$ of the given concave mirror. If the distance of the object is $\mathrm{d}_{1}$ from $\mathrm{C}$ and the distance of the image formed is $d_{2}$ from $C$, the radius of curvature of this mirror is :

  1. $\frac{2 \mathrm{~d}_{1} \mathrm{~d}_{2}}{\mathrm{~d}_{1}-\mathrm{d}_{2}}$

  2. $\frac{2 \mathrm{~d}_{1} \mathrm{~d}_{2}}{\mathrm{~d}_{1}+\mathrm{d}_{2}}$

  3. $\frac{\mathrm{d}_{1} \mathrm{~d}_{2}}{\mathrm{~d}_{1}+\mathrm{d}_{2}}$

  4. $\frac{\mathrm{d}_{1} \mathrm{~d}_{2}}{\mathrm{~d}_{1}-\mathrm{d}_{2}}$


Correct Option: 1

Solution:

Using Newton"s formula

$\left(f+d_{1}\right)\left(f-d_{2}\right)=f^{2}$

$\mathrm{f}^{2}+\mathrm{fd}_{1}-\mathrm{fd}_{2}-\mathrm{d}_{1} \mathrm{~d}_{2}=\mathrm{f}^{2}$

$\mathrm{f}=\frac{\mathrm{d}_{1} \mathrm{~d}_{2}}{\mathrm{~d}_{1}-\mathrm{d}_{2}}$

$\therefore \mathrm{R}=\frac{2 \mathrm{~d}_{1} \mathrm{~d}_{2}}{\mathrm{~d}_{1}-\mathrm{d}_{2}}$

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