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Question

In a convex lens, what happens to the image if the object is placed between the focal point and twice
the focal length?

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Explanation

**Refraction through prism** deviation δ = i+e-A, minimum when i=e, symmetrical path, r₁=r₂=A/2, n = sin[(A+δ_m)/2]/sin(A/2). In water n_rel = n_prism/n_water =1.5/1.33=1.128, so δ_m reduces because relative index lower, new δ_m from formula with n_rel. For a convex lens, when the object is between the focal point (F) and twice the focal length (2F), the image is real, inverted, and magnified. It forms beyond 2F on the opposite side, as the rays converge after refraction. Substituting values gives Real, inverted, and magnified, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror),

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