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Geometrical Optics question

2022 · 27 Jun · Shift 2 · Q60
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  5. /2022 · 27 Jun · Shift 2 · Q60

Geometrical Optics question

2022 · 27 Jun · Shift 2 · Q60

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
A convex lens has power P. It is cut into two halves along its principal axis. Further one piece (out of the two halves) is cut into two halves perpendicular to the principal axis (as shown in figures). Choose the incorrect option for the reported pieces. JEE Main 2022 (Online) 27th June Evening Shift Physics - Geometrical Optics Question 120 English
  1. A
    Power of L1=P2{L_1} = {P \over 2}L1​=2P​
  2. B
    Power of L2=P2{L_2} = {P \over 2}L2​=2P​
  3. C
    Power of L3=P2{L_3} = {P \over 2}L3​=2P​
  4. D
    Power of L1 = P
View written solutionFree

Correct answer: A

  1. Use lens-maker idea for power

For a thin lens in air,

\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$$ So the power depends on the **curvatures of the refracting surfaces**, not on the aperture/size of the lens. --- 2. **First cut: along the principal axis** Cutting a convex lens into two parts **along the principal axis** only reduces its aperture; it does **not** change the radii of curvature of the two refracting surfaces for each piece. Hence each half behaves like a lens of the **same focal length** and therefore the **same power** as the original lens. So for piece $L_1$, $$P_{L_1}=P.$$ Therefore option **D** is correct. Also, option **A** saying $$P_{L_1}=\frac P2$$ is incorrect. --- 3. **Second cut: perpendicular to the principal axis** Now one of the above halves is further cut into two parts **perpendicular to the principal axis**. This means the lens is split into front and back portions, so each resulting piece has one curved surface and one plane surface. Each such piece is a **plano-convex lens**. For a symmetric original biconvex lens, let the original radii be $R_1=R$ and $R_2=-R$. Then original power is $$P=(\mu-1)\left(\frac{1}{R}-\frac{1}{-R}\right)=2(\mu-1)\frac{1}{R}.$$ So each plano-convex half has power $$P'=(\mu-1)\left(\frac{1}{R}-0\right)=(\mu-1)\frac{1}{R}=\frac P2.$$ Thus the two pieces obtained by this cut, namely $L_2$ and $L_3$, each have power $$P_{L_2}=\frac P2, \qquad P_{L_3}=\frac P2.$$ So options **B** and **C** are correct. --- 4. **Check all options** - **A:** $P_{L_1}=\dfrac P2$ → **Incorrect** - **B:** $P_{L_2}=\dfrac P2$ → Correct - **C:** $P_{L_3}=\dfrac P2$ → Correct - **D:** $P_{L_1}=P$ → Correct Hence the **incorrect option** is: $$\boxed{A}$$ --- 5. **Comparison with stored answer** Stored correct answer: **A** My derived answer: **A** They match.
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