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#visual perception

2 public questions tagged with this topic.

Refraction at Spherical Surfaces and Lens Maker's Formula

Why does the apparent depth of an object in a denser medium appear less than its real depth when viewed from air?

**Spherical refracting surface** power P = (n₂-n₁)/R, lens power sum of two surfaces. Lens maker derivation combines two refractions, sign of R₂ negative for second surface convex opposite direction, yielding 1/f positive for convex lens. When light travels from a denser medium (e.g., water) to a rarer medium (air), it bends away from the normal. This refraction makes the rays appear to diverge from a point closer to the surface than the actual object, reducing the apparent depth compared to the real depth. Substituting values gives Due to refraction bending rays away from the normal, which ma

Ref: NCERT > Physics Book > Ray Optics > Refraction at Spherical Surfaces and Lens Maker's Formula

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Refraction Through Prism and Minimum Deviation

An object at a depth of \( 19.95 \, \text{cm} \) in water (\( n = 1.33 \)) is viewed normally. What is the apparent dept

**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. Apparent depth = (real depth/n) . Real depth = 19.95 cm , n = 1.33 . Apparent depth = (19.95/1.33) ≈ 15 cm . Substituting values gives 15 cm, 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), confirming sign conventions and refraction principles.

Ref: NCERT > Physics Book > Ray Optics > Refraction Through Prism and Minimum Deviation

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