Question

D Question 15 1 pts A4 mm high object is placed 5 cm in front of a concave mirror with radius of curvature 20 cm. Questions 13-15 refer to this situation. Time Attem 1 Hou The image will be: Virtual and upright Real and inverted Real and upright Virtual and Inverted

Answer 1

solution 15 Given : concave minor .5cm hu= height of object = 4 mm object distance Ro radius of R=2f focal length (f) = curvature= -20cm Minor formula =-10cm ㅗ u ㅗ f t + 1/2 = 7/10 ch = . 5 2-1 lo to 10 lo 10 cm the is positive ne virtual image. image will be bebrind the mirro magnification (m) 2 = 2 4 -5 me is positive OR upright the image will be errect Ans - virtual and upright

solution - 16 Given: convex mimor. ho= 4 mm u = -9 cm R=2f = 12 cm fa 6 cm + + + 1 = it [ ++ 6 - 9 3+2 = Llog <H- 1 + + L g 18 IS 3.6 cm cm Ans: The image distance between o and 4 cm Solution - 17 hi magnification m = -1 u ho - 3.6 0.4 = hi - 9 ho = height of image (hi) = 0, 4x ho 0, 4 x 4 -1.6mm Ans! The magnitude of the image height will be Less than 2 mm. solution - 18 since v is positive, the image the image will be behind ie virtual image. the mimor it is erect magnification em) is positive means or upright. Ans : The image will will be virtual & upright