RayII 1 A converging lens has a focal length f 20cm when it is in air The lens is made of glass with index of refraction nglass 1 6 When the lens is placed in water nwater 1 33 the focal length of the lens is A Unchanged B greater f 20 cm C smaller f 20cm but still positive D negative Answer greater f 20cm The rays of light are bent when they pass from the medium air or water into the glass according to Snell s law The amount of bending the change in the angle depends on the change in the index of refraction n Bigger change in n means bigger change in angle that is more bending If the lens is immersed in a fluid with the same n as the glass then there is no refraction n 0 0 When the lens is in air the change in n is n 1 6 1 0 6 When the lens is in water the change in n is n 1 6 1 33 0 27 Smaller n in the water means less bending a longer focal length f rays in water less bending of ray longer f rays in air RayII 2 Two point sources of light are imaged onto a screen by a converging lens as shown The images are labeled 1 and 2 A mask is used to cover up the left half of the lens as shown What happens to the images on the screen when the mask is inserted over the left half the lens 1 screen 2 lens mask A Image 1 vanishes B Image 2 vanishes C Something else happens Answer Something else happens Both images dim somewhat but neither disappears Rays from each source cover the entire lens When half the lens is covered half the rays from each source are blocked but the other half get through producing a dimmed image 1 screen 2 lens mask RayII 3 A bundle of parallel rays approaches the eye and some of the rays enter the eye s pupil as shown below No other rays enter the eye What does the eye see Eye A A single point of light surrounded by blackness B A uniformly illuminated wall of light like a white wall C Many scattered points of light like stars in the night sky D None of these Answer A single point of light surrounded by blackness A point source at infinity makes a parallel bundle of rays The lens of the eye focuses these rays onto a point on the retina and the brain perceives a point of light RayII 4 A converging lens focuses the light from a nearby point source onto an image as shown The focal point of a lens is the point on the optic axis one focal length f from the lens Where is the focal point of this lens A between the lens and the image B at the image C further from the lens than the image image A C B object same angle Answer the focal point is between the image and the lens The focal point is where parallel rays along the optic axis would focus A ray hitting the edge of the lens will be bent by same angle regardless of the incoming angle So when the object moves from infinity toward the lens the image moves away from the lens but the focal point remains stationary by definition RayII 5 An object is placed is placed near a diverging lens but the object is further from the lens than the absolute value of the focal length of the lens The image formed is A Real B Virtual C there is no image The magnitude of the image distance compared to the object distance is A smaller B greater object image optic axis Answers The image is virtual and the image distance is smaller than the object distance The only way to understand this is to draw a ray diagram You get the same answers whether or not the object distance is further from the lens than the focal point RayII 6 An object is placed closer to a magnifying glass than the focal length image virtual object di do What are the signs of the focal length f the object distance do and the image distance di A f 0 do 0 di 0 B f 0 do 0 di 0 C f 0 do 0 di 0 D f 0 do 0 di 0 If do 5cm di 15 cm and the object height ho is 1cm what is the image height hi A 2cm B 3cm C 4cm D None of these Answers f positive for converging lenses Object distance positive l 0 for real objects always the case in this course Image distance negative l 0 since image is virtual and on left of lens Image height is 3cm hi d 5 i 3 ho d o 15 RayII 7 A person who is near sighted or myopic cannot focus on faraway objects objects at infinity because the curvature of the person s eye lens is too great This causes parallel rays from a distant point source to bend too much and focus in front of the retina The person sees a fuzzy patch of light rather than a sharp point f too short This person needs eyeglasses with lenses that are A converging B diverging C either converging or diverging depending on much correction is needed Answer diverging lens is needed To be diverging a lens must be thinner in the middle Eyeglasses to correct myopia are shaped like The job of the eyeglasses is to take the real image at l and create a virtual image close enough so the myopic eye can focus on it In my case my myopia is so bad that I can focus on objects no further than 11 cm from my eye So for my glasses when the object is at infinity l the virtual image must be 11 cm from my eye which means it must be about 9 5cm from the glasses l 9 5 cm 0 095 m The lens equation says 1 1 1 1 1 1 10 5 diopters L L f 0 095m f Diopters 1 f where f is in meters Optometrists always indicate lens power in diopters 9 5cm 11 0 cm RayII 9 An astronomical refracting telescope has 3 4 of its objective lens covered with a mask The observer reports that compared to the image with no mask the image is A unchanged B 3 4 covered C the same image but 1 4 as bright D the same image but 1 16 as bright E None of these Answer The image is 1 4 as bright but is otherwise unchanged Actually there will also be some loss of resolution but this will not be very apparent to the viewer unless the magnification is very high RayII 10 What s wrong with that scene in 2001 where Astronaut Dave Bowman is piloting his pod A Astronaut Dave Bowman can t read the screens on his console …
View Full Document
Unlocking...