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Geometrical Optics question

2021 · Shift 1 · Q41
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Geometrical Optics question

2021 · Shift 1 · Q41

JEE AdvancedPhysicsGeometrical OpticsMCQ+3 / −1
An extended object is placed at point O, 10 cm in front of a convex lens L1 and a concave lens L2 is placed 10 cm behind it, as shown in the figure. The radii of curvature of all the curved surfaces in both the lenses ae 20 cm. The refractive index of both the lenses is 1.5. The total magnification of this lens system is JEE Advanced 2021 Paper 1 Online Physics - Geometrical Optics Question 40 English
  1. A
    0.4
  2. B
    0.8
  3. C
    1.3
  4. D
    1.6
View written solutionFree

Correct answer: B

  1. Find focal lengths of the lenses

Using lens maker formula for a thin lens in air: 1f=(μ−1)(1R1−1R2)\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)f1​=(μ−1)(R1​1​−R2​1​) with μ=1.5\mu=1.5μ=1.5.

Since all curved surfaces have radius magnitude 20 cm20\text{ cm}20 cm:

  • For the convex lens L1L_1L1​ (symmetric biconvex): R1=+20 cm,R2=−20 cmR_1=+20\text{ cm},\quad R_2=-20\text{ cm}R1​=+20 cm,R2​=−20 cm So,
=\frac{1}{2}\cdot\frac{2}{20}=\frac{1}{20}$$ Hence, $$f_1=+20\text{ cm}$$ - For the **concave lens $L_2$** (symmetric biconcave): $$R_1=-20\text{ cm},\quad R_2=+20\text{ cm}$$ So, $$\frac{1}{f_2}=(1.5-1)\left(\frac{-1}{20}-\frac{1}{20}\right) =\frac{1}{2}\cdot\left(-\frac{2}{20}\right)=-\frac{1}{20}$$ Hence, $$f_2=-20\text{ cm}$$ --- 2. **Image formed by the first lens $L_1$** Object is $10\text{ cm}$ in front of $L_1$. Using sign convention: $$u_1=-10\text{ cm},\quad f_1=+20\text{ cm}$$ Lens formula: $$\frac{1}{f} = \frac{1}{v}-\frac{1}{u}$$ So, $$\frac{1}{20}=\frac{1}{v_1}-\left(-\frac{1}{10}\right)=\frac{1}{v_1}+\frac{1}{10}$$ $$\frac{1}{v_1}=\frac{1}{20}-\frac{1}{10}=-\frac{1}{20}$$ Thus, $$v_1=-20\text{ cm}$$ So the first lens forms a **virtual image** $20\text{ cm}$ to the left of $L_1$. Magnification by first lens: $$m_1=\frac{v_1}{u_1}=\frac{-20}{-10}=2$$ --- 3. **Object for the second lens $L_2$** $L_2$ is placed $10\text{ cm}$ to the right of $L_1$. The image formed by $L_1$ is $20\text{ cm}$ left of $L_1$. Therefore, relative to $L_2$, this point is: $$20+10=30\text{ cm}$$ left of $L_2$. Hence for $L_2$, $$u_2=-30\text{ cm},\quad f_2=-20\text{ cm}$$ Apply lens formula: $$\frac{1}{f_2}=\frac{1}{v_2}-\frac{1}{u_2}$$ $$-\frac{1}{20}=\frac{1}{v_2}-\left(-\frac{1}{30}\right)=\frac{1}{v_2}+\frac{1}{30}$$ $$\frac{1}{v_2}=-\frac{1}{20}-\frac{1}{30}=-\frac{5}{60}=-\frac{1}{12}$$ So, $$v_2=-12\text{ cm}$$ Magnification by second lens: $$m_2=\frac{v_2}{u_2}=\frac{-12}{-30}=0.4$$ --- 4. **Total magnification** $$m=m_1m_2=2\times 0.4=0.8$$ --- 5. **Check with options** The total magnification is: $$\boxed{0.8}$$ So the correct option is **B**.
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