This gives an image distance of -6.67 cm. Re-arranging the mirror equation to solve for the image distance gives:ĭ i = d o f / (d o - f) = (12) (-15) / (12 -15) The focal ray heads toward F, and reflects off the mirror parallel to the principal axis.Įxtending all three reflected rays back gets you the point where they cross that's where the tip of the image is. The chief ray heads toward C, and comes straight back. The parallel ray goes from the tip of the object, parallel to the principal axis, and reflects off the mirror so that when you extend the reflected ray back it passes through the focal point. On the diagram, sketch the ray diagram showing the position of the image. (b) An object is placed 12 cm from a convex mirror with a focal length of -15 cm. When real images are involved, the object and image are interchangable. If you draw a ray diagram for the first case, the diagram in the second case will be identical, with the directions of the rays all reversed. The image will be where the object was originally, 20 cm from the mirror. Where will the image be now? Briefly explain your reasoning. The image position is marked, and the object is now moved and placed there. (a) An object is placed 20 cm from a concave mirror, forming a real image. Without the second polarizer to rotate the polarization direction, none of the vertically-polarized light gets through the third polarizer, which passes light polarized in the horizontal direction. Zero.Light passed by the first polarizer is polarized along the vertical direction. What is the intensity of light transmitted by the two (i.e., first and third) polarizers now? (d) Suppose the second polarizer is removed. It's the angle between that polarization direction and the polarization axis of the third polarizer that's needed. This is because the second polarizer produces light polarized at 30 degrees to the vertical. The angle that goes in is the difference between the angles for the second and third polarizers, 60 degrees. What is the intensity of the light transmitted by the third polarizer?Īgain, apply Malus' law. (c) The light transmitted by the second polarizer now encounters a third polarizer, which has its transmission axis oriented at 90 degrees relative to the first polarizer. What is the intensity of the light transmitted by the second polarizer?Īpplying Malus' law, the intensity after going through the second polarizer is: (b) The light now passes through a second polarizer, which has its transmission axis rotated by 30 degrees relative to the axis of the first polarizer. Therefore, the transmitted intensity is 500 W / m 2. When unpolarized light passes through a polarizer, the intensity is cut in half. What is the intensity of the light transmitted by the first polarizer? (a) Unpolarized light, with an intensity of 1000 W/m 2 is incident on the first polarizer. This problem deals with light passing through a succession of polarizers as indicated in the sketch: The easiest one to use here is the motional emf equation: Now we can apply an induced emf equation. (c) What is the value of the constant velocity v of the loop? Fortunately, it's easy because we know both the current and resistance in the loop, so we can apply the familiar equation from circuits: (b) What is the induced emf that is generated in the loop and causes the current I to flow?Īlthough this is an induced emf problem, there isn't enough information here to solve for the induced emf using an induced emf equation. The first picture, with current counter-clockwise, is correct. Current flowing counter-clockwise around the loop generates a magentic field out of the page (the right-hand rule can be used to determine this - if you curl your fingers on your right hand in the direction of the current, your thumb points in the direction of the field produced by that current). Lenz's law says that if the magnetic flux through the loop is increasing into the page, the loop will set up a flux out of the page to try to cancel the change. The loop is moving into the field, which points into the page. (a) Which way will the induced current flow in the loop? An induced current of I = 0.043 amperes is flowing around the loop. The magnetic field is perpendicular to the plane of the loop and into the page as shown in the sketch. A square loop of wire with sides of length L = 0.17 m and total resistance R = 50 ohms is moving with constant velocity v into a region of constant magnetic field B = 0.76 T.

Initial light intensity equation value