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JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
An object is placed beyond the centre of curvature \({C}\) of the given concave mirror. If the distance of the object is \({d}_{1}\) from \({C}\) and the distance of the image formed is \({d}_{2}\) from \({C}\), the radius of curvature of this mirror is
- A \(\frac{2 {d}_{1} {d}_{2}}{{d}_{1}-{d}_{2}}\)
- B \(\frac{2 {d}_{1} {d}_{2}}{{d}_{1}+{d}_{2}}\)
- C \(\frac{{d}_{1} {d}_{2}}{{d}_{1}+{d}_{2}}\)
- D \(\frac{{d}_{1} {d}_{2}}{{d}_{1}-{d}_{2}}\)
Answer & Solution
Correct Answer
(A) \(\frac{2 {d}_{1} {d}_{2}}{{d}_{1}-{d}_{2}}\)
Step-by-step Solution
Detailed explanation
Using Newton"s formula \(\left({f}+{d}_{1}\right)\left({f}-{d}_{2}\right)={f}^{2}\) \({f}^{2}+{fd}_{1}-{fd}_{2}-{d}_{1} {d}_{2}={f}^{2}\) \({f}=\frac{{d}_{1} {d}_{2}}{{d}_{1}-{d}_{2}}\) \(\therefore {R}=\frac{2 {d}_{1} {d}_{2}}{{d}_{1}-{d}_{2}}\)
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