Ch06 Geometrical Optics

PHYSICS 2ND PAPER CHAPTER 6 Geometrical Optics Question 1 [Din. B. 2017] An object 𝑃𝑄is placed at a distance of 50 cm in front of a biconvex glass lens as shown in the diagram. The radii of curvature of the surfaces are 𝑟1 = 30 cm and 𝑟2 = 60 cm. The refractive index of glass is 𝜇𝑔= 1.5 and the refractive index of water is 𝜇𝑤= 1.33. 𝑂 𝑄 𝑃 50 cm 𝜇𝑔= 1.5 𝜇𝑤= 1.33 𝑟1 = 30 cm 𝑟2 = 60 cm (a) Determine the focal length of the lens in air from the stem. (b) If the lens is placed in air and water alternately, determine the nature and position of the images formed, and explain mathematically. Solution (a) Here, Radius of first surface 𝑅1 = 30 cm Radius of second surface 𝑅2 = −60 cm Refractive index of glass 𝜇𝑔= 1.5 By the lens maker’s formula in air: 1 𝑓𝑎 = (𝜇𝑔−1)  1 𝑅1 −1 𝑅2  = (1.5 −1)  1 30 + 1 60  = 0.5 × 3 60 = 1 40 cm−1 𝑓𝑎= 40 cm 𝑓𝑎= 40 cm (b) Sign convention: distances are measured from the lens. They are positive in the direction of the incident ahbab.dev 1

light and negative against it. The lens formula is 1 𝑣−1 𝑢= 1 𝑓and the magnification is 𝑚= 𝑣 𝑢. The object is on the left, so 𝑢= −50 cm. Case 1: lens in air ( 𝑓𝑎= 40 cm). 1 𝑣𝑎 = 1 𝑓𝑎 + 1 𝑢 = 1 40 −1 50 = 5 −4 200 = 1 200 cm−1 𝑣𝑎= 200 cm 𝑚𝑎= 𝑣𝑎 𝑢= 200 −50 = −4 𝑣𝑎is positive and 𝑚𝑎is negative. So the image is real, inverted and 4 times magnified. It forms 200 cm behind the lens. Case 2: lens in water. The refractive index of glass relative to water is 𝑤𝜇𝑔= 𝜇𝑔 𝜇𝑤 = 1.5 1.33 = 1.1278 The focal length in water is 1 𝑓𝑤 = (𝑤𝜇𝑔−1)  1 𝑅1 −1 𝑅2  = (1.1278 −1) × 1 20 = 0.006391 cm−1 𝑓𝑤= 156.47 cm Now the object (𝑢= −50 cm) lies inside the focal length. So 1 𝑣𝑤 = 1 𝑓𝑤 + 1 𝑢 = 0.006391 −0.020000 = −0.013609 cm−1 𝑣𝑤= −73.48 cm 𝑚𝑤= 𝑣𝑤 𝑢= −73.48 −50 = +1.47 𝑣𝑤is negative and 𝑚𝑤is positive. So the image is virtual, erect and 1.47 times magnified. It forms 73.48 cm from the lens, on the same side as the object. 𝐹 𝐹′ 𝑃 image (i) Lens in air: real, inverted image ahbab.dev 2

𝐹 𝐹′ 𝑃 image (ii) Lens in water: virtual, erect image Conclusion: in air the image is real and inverted at 200 cm behind the lens. In water the focal length becomes much longer, so the image becomes virtual and erect, 73.48 cm in front of the lens. 𝑣𝑎= 200 cm (real, inverted), 𝑣𝑤= −73.48 cm (virtual, erect) ahbab.dev 3