Refraction at Plane Surfaces and Snell's Law

Refraction at Plane Surfaces and Snell's Law

What is the critical angle for a glass (\( n = 1.5 \)) to water (\( n = 1.33 \)) interface?

**Refractive index** n = c/v, water 1.33 means light 1.33 times slower than vacuum. Passing from water to air at 49°, n₁ sinθ₁ =1.33×sin49°≈1.33×0.755=1.004>1, so sinθ₂>1 impossible, total internal reflection occurs, no refraction. Critical angle: sin i_c = (n₂/n₁) . Glass ( n₁ = 1.5 ), water ( n₂ = 1.33 ). sin i_c = (1.33/1.5) ≈ 0.887 . i_c = sin⁻¹(0.887) ≈ 62.5° . Substituting values gives 62°, 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.

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Refraction at Plane Surfaces and Snell's Law

A concave mirror of focal length \( 6 \, \text{cm} \) has an object placed \( 12 \, \text{cm} \) from it. What is the im

**Refraction at plane surface** changes direction due to speed change v = c/n, n = c/v. For water n=1.33 to air, n₁>n₂, ray bends away, if θ₁ > C, sinθ₂>1 impossible, TIR occurs. Snell's law quantitative prediction of θ₂ from θ₁ and indices. Focal length: f = -6 cm (concave mirror). Object distance: u = -12 cm . Mirror equation: (1/v) + (1/u) = (1/f) . (1/v) + (1/-12) = (1/-6) ⇒ (1/v) = (1/-6) + (1/12) = (-2 + 1/12) = (-1/12) . v = -12 cm (real image). Substituting values gives 12 cm, which matches expected image position and magnification from mirror/lens formula

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Refraction at Plane Surfaces and Snell's Law

A converging beam of light meets a concave lens of focal length \( 25 \, \text{cm} \) at \( 10 \, \text{cm} \) before th

**Refraction at plane surface** changes direction due to speed change v = c/n, n = c/v. For water n=1.33 to air, n₁>n₂, ray bends away, if θ₁ > C, sinθ₂>1 impossible, TIR occurs. Snell's law quantitative prediction of θ₂ from θ₁ and indices. u = -10 cm (virtual object), f = -25 cm . (1/v) - (1/-10) = (1/-25) ⇒ (1/v) + (1/10) = (1/-25) . (1/v) = (1/-25) - (1/10) = (-2 - 5/50) = (-7/50) . v = -(50/7) ≈ -7.14 cm (7.14 cm to the left). Substituting values gives 7.14 cm, which matches expected image position and magnification from mirror/lens formula 1/f

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Refraction at Plane Surfaces and Snell's Law

A lens has a power of \( +4 \, \text{D} \). What is its focal length in centimeters?

**Snell's law** n₁ sinθ₁ = n₂ sinθ₂ describes refraction at plane interface, n refractive index, θ angle with normal. When light goes from denser n=1.52 glass to rarer air n=1, sinθ₂ = (n₁/n₂) sinθ₁ > sinθ₁, bending away from normal, enabling total internal reflection beyond critical angle. Power: P = (1/f) (in meters). P = +4 D ⇒ 4 = (1/f) ⇒ f = (1/4) = 0.25 m = 25 cm . Substituting values gives 25 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.

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Refraction at Plane Surfaces and Snell's Law

In a convex mirror, what is the significance of the focal point being behind the mirror?

**Refraction at plane surface** changes direction due to speed change v = c/n, n = c/v. For water n=1.33 to air, n₁>n₂, ray bends away, if θ₁ > C, sinθ₂>1 impossible, TIR occurs. Snell's law quantitative prediction of θ₂ from θ₁ and indices. The focal point of a convex mirror is virtual and behind the mirror because reflected rays diverge and appear to originate from this point when traced backward. This indicates the mirror’s diverging nature, ensuring all images are virtual, erect, and diminished. Substituting values gives It is the point where diverging rays appear to originate, which matches expected image position and magnification from

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Refraction at Plane Surfaces and Snell's Law

A concave lens of focal length \( 25 \, \text{cm} \) has an object placed \( 50 \, \text{cm} \) from it. What is the ima

**Refraction at plane surface** changes direction due to speed change v = c/n, n = c/v. For water n=1.33 to air, n₁>n₂, ray bends away, if θ₁ > C, sinθ₂>1 impossible, TIR occurs. Snell's law quantitative prediction of θ₂ from θ₁ and indices. Focal length: f = -25 cm (concave lens). Object distance: u = -50 cm . Lens formula: (1/v) - (1/u) = (1/f) . (1/v) - (1/-50) = (1/-25) ⇒ (1/v) + (1/50) = (1/-25) ⇒ (1/v) = (1/-25) - (1/50) = (-2 - 1/50) = (-3/50) . v = -(50/3) ≈ -16.67 cm (virtual image). Substituting values gives 16.7 cm, which matches

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Refraction at Plane Surfaces and Snell's Law

A ray of light passes from glass (\( n = 1.62 \)) to air at an angle of incidence equal to the critical angle. What is t

**Refractive index** n = c/v, water 1.33 means light 1.33 times slower than vacuum. Passing from water to air at 49°, n₁ sinθ₁ =1.33×sin49°≈1.33×0.755=1.004>1, so sinθ₂>1 impossible, total internal reflection occurs, no refraction. Critical angle: sin i_c = (n₂/n₁) = (1/1.62) ≈ 0.617 . i_c = sin⁻¹(0.617) ≈ 38.1° . At critical angle, angle of refraction = 90° . Substituting values gives 90°, 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.

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Refraction at Plane Surfaces and Snell's Law

In a convex lens, why does the image transition from virtual to real as the object moves from inside to outside the foca

**Refractive index** n = c/v, water 1.33 means light 1.33 times slower than vacuum. Passing from water to air at 49°, n₁ sinθ₁ =1.33×sin49°≈1.33×0.755=1.004>1, so sinθ₂>1 impossible, total internal reflection occurs, no refraction. Inside the focal point, a convex lens diverges rays, forming a virtual image on the same side. Beyond the focal point, the lens converges rays to a point on the opposite side, forming a real image. This transition occurs as the object crosses the focal point, changing the ray behavior. Substituting values gives Due to change from divergence to convergence, which matches expected image position and magnification from mirror/lens formula 1/f =

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Refraction at Plane Surfaces and Snell's Law

A ray of light passes from glass (\( n = 1.5 \)) to air at an angle of incidence of \( 40^\circ \). What is the angle of

**Refraction at plane surface** changes direction due to speed change v = c/n, n = c/v. For water n=1.33 to air, n₁>n₂, ray bends away, if θ₁ > C, sinθ₂>1 impossible, TIR occurs. Snell's law quantitative prediction of θ₂ from θ₁ and indices. Snell’s law: n₁ sin i = n₂ sin r . Glass ( n₁ = 1.5 ), air ( n₂ = 1 ), i = 40° . 1.5 × sin 40° = 1 × sin r . sin 40° ≈ 0.643 ⇒ 1.5 × 0.643 ≈ 0.964 ⇒ sin r = 0.964 . r = sin⁻¹(0.964) ≈ 74.6° . Critical angle: sin i_c

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Refraction at Plane Surfaces and Snell's Law

An object is at a depth of \( 16 \, \text{cm} \) in water (\( n = 1.33 \)). What is the apparent depth?

**Refractive index** n = c/v, water 1.33 means light 1.33 times slower than vacuum. Passing from water to air at 49°, n₁ sinθ₁ =1.33×sin49°≈1.33×0.755=1.004>1, so sinθ₂>1 impossible, total internal reflection occurs, no refraction. Apparent depth = (real depth/n) . Real depth = 16 cm , n = 1.33 . Apparent depth = (16/1.33) ≈ 12.03 cm . Substituting values gives 12 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.

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Refraction at Plane Surfaces and Snell's Law

What characteristic of a convex mirror makes it suitable for use as a rear-view mirror in vehicles?

**Snell's law** n₁ sinθ₁ = n₂ sinθ₂ describes refraction at plane interface, n refractive index, θ angle with normal. When light goes from denser n=1.52 glass to rarer air n=1, sinθ₂ = (n₁/n₂) sinθ₁ > sinθ₁, bending away from normal, enabling total internal reflection beyond critical angle. A convex mirror always produces a virtual, erect, and diminished image, regardless of the object’s position. This provides a wide field of view, allowing drivers to see a larger area behind the vehicle, making it ideal for rear-view mirrors despite the reduced image size. Substituting values gives Provides a wide field of view with a diminished image, which

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Refraction at Plane Surfaces and Snell's Law

A convex mirror of focal length \( 18 \, \text{cm} \) produces an image \( 6 \, \text{cm} \) behind the mirror. What is

**Refractive index** n = c/v, water 1.33 means light 1.33 times slower than vacuum. Passing from water to air at 49°, n₁ sinθ₁ =1.33×sin49°≈1.33×0.755=1.004>1, so sinθ₂>1 impossible, total internal reflection occurs, no refraction. Focal length: f = 18 cm (convex mirror). Image distance: v = 6 cm (virtual image). Mirror equation: (1/v) + (1/u) = (1/f) . (1/6) + (1/u) = (1/18) ⇒ (1/u) = (1/18) - (1/6) = (1 - 3/18) = (-2/18) = (-1/9) . u = -9 cm . Substituting values gives 12 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f

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