Question

Question 9 40 pts A +2.5 cm focal length lens is used to project the image of a slide onto a screen. The lens is 3.0 m from the screen and the screen is 1.5 m wide and 0.5 m high. Part A) 20 pts How far should the slide be from the lens to produce a crisp image on the screen? Part B) 20 pts What are the dimensions of the slide (i.e. width and height) that would make an image that fills the entire screen? Upload

Answer 1

Again, Imag width mm Object width Object width a (120.97)" 0.0102 cm An 5 : The height and width of slide must be 0.4133m and 0.0102 em respectively,

A focal length of- +2.5cm = +0.025 m Smage distance = v=- 3.0 mm Let the object distance from the lens be u. Puen *** = $ * t = 174 17 0.025 + 3.0 u= 0.0248 m = 2.48 m Amb: The slide should be 2.48 em apart from the lens. ation =m= 3.0 0.0248 = 120.97 Magnification Image height Now, 120.97 Object height object height = 120.97 so 0.5 0.4133 em

A focal length of- +2.5cm = +0.025 m Smage distance = v=- 3.0 mm Let the object distance from the lens be u. Puen *** = $ * t = 174 17 0.025 + 3.0 u= 0.0248 m = 2.48 m Amb: The slide should be 2.48 em apart from the lens. ation =m= 3.0 0.0248 = 120.97 Magnification Image height Now, 120.97 Object height object height = 120.97 so 0.5 0.4133 em

Again, Imag width mm Object width Object width a (120.97)" 0.0102 cm An 5 : The height and width of slide must be 0.4133m and 0.0102 em respectively,

A focal length of- +2.5cm = +0.025 m Smage distance = v=- 3.0 mm Let the object distance from the lens be u. Puen *** = $ * t = 174 17 0.025 + 3.0 u= 0.0248 m = 2.48 m Amb: The slide should be 2.48 em apart from the lens. ation =m= 3.0 0.0248 = 120.97 Magnification Image height Now, 120.97 Object height object height = 120.97 so 0.5 0.4133 em