Magnification (m) = h'/h, thus h' = m * h = 3 * 2 = 6 cm.

A virtual image is formed by a concave lens when the object is placed in front of it.

Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1 (air), n2 = 1.5, i = 30 degrees. Solving gives r = 19.2 degrees.

Since the angle of incidence is less than the critical angle (θc = sin⁻¹(1/1.5) ≈ 41.8°), it will not undergo total internal reflection.

Using Snell's law, n1 * sin(theta1) = n2 * sin(theta2). Here, n1 = 1 (air), n2 = 1.5 (glass), and theta1 = 30 degrees. Thus, sin(theta2) = (1 * sin(30))/1.5 = 0.333, giving theta2 ≈ 19.1 degrees.

Critical angle θc = sin⁻¹(1.00/2.42) ≈ 24.4°.

When a light ray passes through the center of curvature, it strikes the mirror perpendicularly, resulting in an angle of reflection of 0 degrees.

Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, sin(r) = (1 * sin(45)) / 1.5 = 0.471, giving r ≈ 28 degrees.

Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1 (air), n2 = 1.5 (glass), i = 60 degrees. Solving gives r = 40 degrees.

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1 (air), θ1 = 30 degrees, n2 = 1.5. Thus, sin(θ2) = (1 * sin(30))/1.5 = 1/3, giving θ2 ≈ 20 degrees.

According to the law of reflection, the angle of reflection equals the angle of incidence, so it is 45 degrees.

The critical angle for this scenario is approximately 38.7°. Since 50° is greater than the critical angle, total internal reflection occurs.

Since 50° is greater than the critical angle (θc ≈ 41.8°), total internal reflection occurs.

Calculate critical angle: θc = sin^(-1)(1/n) = sin^(-1)(1/1.6) ≈ 38.7°. Since 50° > 38.7°, total internal reflection occurs.

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 2.0, θ1 = 45°, and n2 = 1.0 (air). Thus, 2.0 * sin(45°) = 1.0 * sin(θ2) leads to sin(θ2) = √2, which gives θ2 = 90°.

Using sin(θc) = n2/n1, where n1 = 2.42 (diamond) and n2 = 1.0 (air), we find sin(θc) = 1.0/2.42, leading to θc ≈ 24.4°.

Using Snell's law, sin(θc) = n2/n1 = 1.00/2.42. Thus, θc ≈ 24.4°.

Since sin(30°) < sin(θc) where θc = sin⁻¹(1/1.5) ≈ 41.8°, it will not undergo total internal reflection.

Since 45° < θc ≈ 48.6°, the ray will partially reflect and refract.

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1.5, θ1 = 30°, n2 = 1.0. Thus, sin(θ2) = (1.5 * sin(30°))/1.0 = 0.75, giving θ2 ≈ 60°.

Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1 (air), i = 30 degrees, n2 = 1.5 (glass). Thus, sin(r) = (1 * sin(30))/1.5 = 0.333, giving r ≈ 18.4 degrees.

Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1 (air), n2 = 1.5 (glass), i = 45 degrees. Solving gives r = 30 degrees.

Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1, n2 = 1.5, i = 30 degrees. Solving gives r ≈ 18.4 degrees.

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, sin(30) = 0.5, so 1 * 0.5 = 1.33 * sin(θ2). Solving gives θ2 ≈ 22.5 degrees.

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1 (air), n2 = 1.33 (water), θ1 = 30 degrees. Thus, sin(θ2) = (1 * sin(30))/1.33 = 0.375. Therefore, θ2 = sin^(-1)(0.375) which is approximately 22 degrees.

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2), we find θ2 = sin^(-1)(sin(30 degrees)/1.33) = 22.5 degrees.

Using Snell's law, n1*sin(θ1) = n2*sin(θ2), we find θ2 = 22.1 degrees.

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1 (air), n2 = 1.33 (water). Solving gives θ2 ≈ 22 degrees.

Using Snell's law: n1 * sin(θ1) = n2 * sin(θ2). Here, sin(30) = 0.5, so sin(θ2) = (1 * 0.5) / 1.33 = 0.375. Thus, θ2 = sin^(-1)(0.375) ≈ 22.5 degrees.

When a ray of light passes through the center of curvature, it strikes the mirror perpendicularly, resulting in an angle of reflection of 0 degrees.