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Geometrical Optics question

2021 · 27 Aug · Shift 1 · Q46
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  5. /2021 · 27 Aug · Shift 1 · Q46

Geometrical Optics question

2021 · 27 Aug · Shift 1 · Q46

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
An object is placed beyond the centre of curvature C of the given concave mirror. If the distance of the object is d1 from C and the distance of the image formed is d2 from C, the radius of curvature of this mirror is :
  1. A
    2d1d2d1−d2{{2{d_1}{d_2}} \over {{d_1} - {d_2}}}d1​−d2​2d1​d2​​
  2. B
    2d1d2d1+d2{{2{d_1}{d_2}} \over {{d_1} + {d_2}}}d1​+d2​2d1​d2​​
  3. C
    d1d2d1+d2{{{d_1}{d_2}} \over {{d_1} + {d_2}}}d1​+d2​d1​d2​​
  4. D
    d1d2d1−d2{{{d_1}{d_2}} \over {{d_1} - {d_2}}}d1​−d2​d1​d2​​
View written solutionFree

Correct answer: A

  1. Set up the geometry

For a concave mirror, let the pole be at PPP, the centre of curvature at CCC, and radius of curvature be R=PCR = PCR=PC.

The object is placed beyond CCC, so if its distance from CCC is d1d_1d1​, then its distance from the pole is u=−(R+d1)u = -(R + d_1)u=−(R+d1​) (using Cartesian sign convention; object is to the left of the mirror).

The image formed by a concave mirror for an object beyond CCC lies between FFF and CCC. If its distance from CCC is d2d_2d2​, then its distance from the pole is v=−(R−d2)v = -(R - d_2)v=−(R−d2​) because the image is also to the left of the mirror and is at a distance R−d2R-d_2R−d2​ from the pole.

Also, for a concave mirror, f=−R2f = -\frac{R}{2}f=−2R​

  1. Apply mirror formula

Mirror formula: 1f=1v+1u\frac{1}{f} = \frac{1}{v} + \frac{1}{u}f1​=v1​+u1​

Substitute: 1−R/2=1−(R−d2)+1−(R+d1)\frac{1}{-R/2} = \frac{1}{-(R-d_2)} + \frac{1}{-(R+d_1)}−R/21​=−(R−d2​)1​+−(R+d1​)1​

So, −2R=−1R−d2−1R+d1-\frac{2}{R} = -\frac{1}{R-d_2} - \frac{1}{R+d_1}−R2​=−R−d2​1​−R+d1​1​

Multiply by −1-1−1: 2R=1R−d2+1R+d1\frac{2}{R} = \frac{1}{R-d_2} + \frac{1}{R+d_1}R2​=R−d2​1​+R+d1​1​

  1. Simplify the right-hand side
= \frac{(R+d_1)+(R-d_2)}{(R-d_2)(R+d_1)} = \frac{2R + d_1 - d_2}{(R-d_2)(R+d_1)}$$ Hence, $$\frac{2}{R} = \frac{2R + d_1 - d_2}{(R-d_2)(R+d_1)}$$ Cross-multiply: $$2(R-d_2)(R+d_1) = R(2R + d_1 - d_2)$$ 4. **Expand and solve for $R$** Left side: $$2(R-d_2)(R+d_1) = 2\left(R^2 + Rd_1 - Rd_2 - d_1d_2\right)$$ $$= 2R^2 + 2R(d_1-d_2) - 2d_1d_2$$ Right side: $$R(2R + d_1 - d_2) = 2R^2 + R(d_1-d_2)$$ Equating: $$2R^2 + 2R(d_1-d_2) - 2d_1d_2 = 2R^2 + R(d_1-d_2)$$ Cancel $2R^2$: $$2R(d_1-d_2) - 2d_1d_2 = R(d_1-d_2)$$ $$R(d_1-d_2) = 2d_1d_2$$ Therefore, $$R = \frac{2d_1d_2}{d_1-d_2}$$ 5. **Match with options** This is exactly $$\frac{2d_1d_2}{d_1-d_2}$$ which corresponds to **Option A**. --- **Final Answer:** $\boxed{\dfrac{2d_1d_2}{d_1-d_2}}$ (Option A)
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