Answer: Option B

Answer: Option A Refer to the diagram (below left) which many are familiar. When parallel rays of light which are parallel to the principal axis enter the lens, the rays bend (refraction), come closer and converge to a point on the principal axis called focal point (F) . The distance from the optical centre (C) to the focal point (F) is the focal length (f) . But what if the parallel rays of light entering the lens are not parallel to the principal axis but at an angle as shown on the diagram (below right)? As you can see, the rays refracted and converge to a point P which is along the focal plane (imaginary vertical line through F and is perpendicular to the principal axis). This is similar to L1 in the question. (Refer to the first section of the video simulation below to reinforce your concept) How about L2 in the question? Light is reversible so you can also treat the light rays entering from the right of the lens L2. The parallel rays of light in L2 are at an angle but there is no ray through the optical centre C. Refer to the video below, as you can see, the parallel rays of light will likewise refract and converge to a point, which is along the focal plane too. Hence the focal point of both lenses L1 and L2 is at F2. So the answer is Option A.

Answer: Option C

Solutions: a) Refraction     b) 48.6o     c) c = 41.8o     e) 70o d) Total internal reflection has occured, as the angle of incidence is greater than critical angle and the light is traveling from a denser medium (glass) towards a less dense medium (air) Click on the video tutorial for explanation and working for part (a) and (b). Click on the video tutorial for explanation and working for part (c) to (e).

Solutions: a) 38.8o      b) 41.8o c) 51.2o . The ray does not emerge as total internal reflection has occurred. The angle of incidence is greater than critical angle and the light is traveling from a denser towards a less dense medium. Click here to view the video tutorial for working and explanation for (a) and (b). Click here to view the video tutorial for working and explanation for (c) and (d).

Jet Suit Flight and Talk by Richard Browning Setting Records Real Ironman Another kind of jet suit by another company

In this post, it shows a free-falling ball from a height of 1.0 m. During the impact, the direction of the force on the ground is downwards and the force on the ground by the ball is greater then the weight. As the ball is free-falling, the only force acting is its weight downwards. Hence a common misconception is to think that the force on the ground during impact is equal to the weight. This is wrong. The normal force (force on the ball by the ground = stopping force on the ball by the ground) is greater than the weight . The force on the ball by the ground is equal and opposite to the force on the ground by the ball . Hence the magnitude of the force on the ground is greater than the weight . Similar concept can be applied if a man jumps off from a height. But in this case, the man’s leg will exert a stopping force over a short distance. That stopping force, once again, is greater than the weight of the man.