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NEET PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSMedium

Question

A biconvex lens of refractive index 1.5, has a radius of curvature of magnitude 20 cm20\text{ cm}. Which one of the following options describes best the image formed of an object of height 2 cm2\text{ cm} placed 30 cm30\text{ cm} from the lens?

A

Virtual, upright, height=0.5 cm0.5\text{ cm}

B

Real, inverted, height=4 cm4\text{ cm}

C

Real, inverted, height=1 cm1\text{ cm}

D

Virtual, upright, height=1 cm1\text{ cm}

Step-by-Step Solution

  1. Focal Length of the Lens (ff): Using the Lens Maker's formula, 1f=(μ−1)(1R1−1R2)\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right). For a biconvex lens, by sign convention, R1=+20 cmR_1 = +20\text{ cm} and R2=−20 cmR_2 = -20\text{ cm}. 1f=(1.5−1)(120−1−20)=0.5×(220)=0.5×0.1=0.05=120\frac{1}{f} = (1.5 - 1)\left(\frac{1}{20} - \frac{1}{-20}\right) = 0.5 \times \left(\frac{2}{20}\right) = 0.5 \times 0.1 = 0.05 = \frac{1}{20}. Thus, f=+20 cmf = +20\text{ cm}.
  2. Image Distance (vv): Using the thin lens formula 1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f}. The object distance u=−30 cmu = -30\text{ cm}. 1v−(1−30)=120  ⟹  1v+130=120\frac{1}{v} - \left(\frac{1}{-30}\right) = \frac{1}{20} \implies \frac{1}{v} + \frac{1}{30} = \frac{1}{20} 1v=120−130=3−260=160\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}. Thus, v=+60 cmv = +60\text{ cm}. The positive sign indicates that the image forms on the opposite side of the lens, meaning it is a Real image.
  3. Magnification (mm) and Image Height (hih_i): Linear magnification is given by m=vum = \frac{v}{u}. m=+60−30=−2m = \frac{+60}{-30} = -2. The negative sign of magnification indicates that the image is inverted. Now, m=hiho  ⟹  hi=m×ho=−2×2 cm=−4 cmm = \frac{h_i}{h_o} \implies h_i = m \times h_o = -2 \times 2\text{ cm} = -4\text{ cm}. The magnitude of the height is 4 cm4\text{ cm}.
  4. Conclusion: The image is real, inverted, and has a height of 4 cm4\text{ cm}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from RAY OPTICS AND OPTICAL INSTRUMENTS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSbiconvexrefractiveradiuscurvaturemagnitude

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