c846aabc20c3.webp" data-orsrc="https://uploads-pracbee-ramanuj.s3.ap-south-1.amazonaws.com/pyq-question-images/batch-001/c846aabc20c3.webp" loading="lazy" style="max-width:100%;height:auto;display:block;margin:0 auto;max-height:40vh;vertical-align:baseline" alt="JEE Main 2025 (Online) 3rd April Evening Shift Physics - Geometrical Optics Question 36 English Explanation">

We apply the refraction formula at a spherical surface: μ₂/v − μ₁/u = (μ₂ − μ₁)/R.

Here:

• μ₁ = 1 (air)
• μ₂ = 1.5 (glass)
• R = 50 cm (radius of curvature)
• v = +200 cm (image distance inside glass)
• u = –x (object distance, negative for a real object in front of the surface)

Substitute these values and solve for x:

$$\begin{aligned} & \frac{\mu_2}{\mathrm{v}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\ & \frac{1.5}{200}-\frac{1}{-\mathrm{x}}=\frac{1.5-1}{50} \\ & \frac{1}{\mathrm{x}}=\frac{1}{100}-\frac{3}{400} \\ & \mathrm{x}=400 \mathrm{~cm} \\ & \mathrm{x}=4 \mathrm{~m} \end{aligned}$$

So, the light source is 4 m from the glass surface.

19:T677,

JEE Main 2025 (Online) 3rd April Evening Shift Physics - Geometrical Optics Question 36 Hindi Explanation

हम गोलीय पृष्ठ पर अपवर्तन का सूत्र लागू करते हैं: μ₂/v − μ₁/u = (μ₂ − μ₁)/R।

यहाँ:

• μ₁ = 1 (वायु)
• μ₂ = 1.5 (काँच)
• R = 50 cm (वक्रता त्रिज्या)
• v = +200 cm (काँच के अंदर प्रतिबिम्

Geometrical Optics

Continue this paper

Similar Geometrical Optics PYQs

More practice from this question