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Let AP be a section of a convex spherical surface, separating two media of refractive indices and . C is the curvature and P is the pole. MN is a small linear object situated on the axis far away from P in the medium of refractive index . The image of the point N is formed at a point N’ on the axis. To find the image of the point M, take a ray MA parallel to the principal axis which after refraction passes through the second focus F. Another ray MC passes through the centre of curvature and is undeviated. The two rays intersect at the point M’ which is the image of M. Therefore, M’N’ is the linear image of the object MN. Δ s MNC and M’N’C are similar. From the equation From equations, (i), (ii) and (iii), From equation (iv) it follows that when v is positive, the image is real and inverted. When v is negative, the image is virtual and erect. Services: - Lateral Transverse Magnification Homework | Lateral Transverse Magnification Homework Help | Lateral Transverse Magnification Homework Help Services | Live Lateral Transverse Magnification Homework Help | Lateral Transverse Magnification Homework Tutors | Online Lateral Transverse Magnification Homework Help | Lateral Transverse Magnification Tutors | Online Lateral Transverse Magnification Tutors | Lateral Transverse Magnification Homework Services | Lateral Transverse Magnification
Let AP be a section of a convex spherical surface, separating two media of refractive indices and . C is the curvature and P is the pole. MN is a small linear object situated on the axis far away from P in the medium of refractive index . The image of the point N is formed at a point N’ on the axis. To find the image of the point M, take a ray MA parallel to the principal axis which after refraction passes through the second focus F. Another ray MC passes through the centre of curvature and is undeviated. The two rays intersect at the point M’ which is the image of M. Therefore, M’N’ is the linear image of the object MN. Δ s MNC and M’N’C are similar. From the equation
From equations, (i), (ii) and (iii), From equation (iv) it follows that when v is positive, the image is real and inverted. When v is negative, the image is virtual and erect.
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