This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Step 1: Determine the image distance for the concave mirror. Given: Object distance, u = -30 cm (negative for a real object placed in front of the mirror) Focal length of concave mirror, f = -20 cm (negative for a concave mirror) The mirror formula is given by: (1)/(f) = (1)/(v) + (1)/(u) Rearrange to solve for the image distance v: (1)/(v) = (1)/(f) - (1)/(u) Substitute the given values: (1)/(v) = (1)/(-20 cm) - (1)/(-30 cm) (1)/(v) = -(1)/(20 cm) + (1)/(30 cm) Find a common denominator (60): (1)/(v) = (-3)/(60 cm) + (2)/(60 cm) (1)/(v) = (-1)/(60 cm) v = -60 cm The negative sign indicates that the image is real and formed in front of the mirror. Step 2: Determine the magnification. The magnification M is given by: M = -(v)/(u) Substitute the values of v and u: M = --60 cm-30 cm M = -2 The negative sign indicates that the image is inverted. The magnitude 2 means the image is twice the size of the object. Step 3: Explain how to draw the ray diagram. To draw the ray diagram for a concave mirror with an object placed at u = 30 cm and f = 20 cm: • Draw a principal axis and place the concave mirror. Mark the focal point F at 20 cm from the mirror and the center of curvature C at 40 cm from the mirror. • Place the object (needle) at 30 cm from the mirror, between F and C. • Draw three principal rays from the top of the object: 1. A ray parallel to the principal axis, which reflects through the focal point F. 2. A ray passing through the focal point F, which reflects parallel to the principal axis. 3. A ray passing through the center of curvature C, which reflects back along the same path. • The point where these reflected rays intersect is the top of the image. The image will be real, inverted, and magnified, located at 60 cm from the mirror (beyond C). Step 4: Calculate the speed of light in the medium. Given: Critical angle, C = 45^ Speed of light in vacuum, c = 3 × 10^8 m/s First, calculate the refractive index n of the medium using the critical angle formula: (C) = (1)/(n) n = (1)/((C)) n = (1)/((45^)) n = (1)/(sqrt(2))2 n = sqrt(2) ≈ 1.414 Next, calculate the speed of light v in the medium using the refractive index formula: n = (c)/(v) v = (c)/(n) v = 3 × 10^8 m/ssqrt(2) v ≈ 3 × 10^8 m/s1.4142 v ≈ 2.121 × 10^8 m/s The image distance is -60 cm, the magnification is -2, and the speed of light in the medium is approximately 2.12 × 10^8 m/s. What's next?