HRWCh33(1).pdf HRWCh33(1).pdf v/pid-3099245-dt-content-rid-1981700 - + Automatic
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HRWCh33(1).pdf HRWCh33(1).pdf v/pid-3099245-dt-content-rid-1981700 - + Automatic Zoom Reading sour Pedottich2.pdf ×-HRNG D Duquesne Un DOR e https://duq.blackboard 32of34 the plane surface, one sees two images of the bubble. How do they arise and where do they appear? from the window determine the na fraction due to th 5cm n=1.50 Figu 2-16 A plano-convex le be made with glass Figure 2-34 Problenn 2-10 radius of curvaturc used in making this 2-17 Calculate the focal le ical surfaces have rad The glass is of index 2-11 A concave mitror forms an image on a screen twice as large as the object. Both object and screen are then moved to produce an image on the screen that is three times the size of the object. If the screen is moved 75 cm in the process how far is the object moved? What is the focal length of the mirror? versions of the lens 2-18 One side of a ish tan made of glass (n = 1 curvature 30 em. A s lens. Where does the lens? What is its mag 2-12 A sphere 5 cm in diameter has a small scratch on its surface. When the scratch is viewed through the glass from a position directly opposite, where does the scratch appear and what is its magnification? Ascume n1.50 for the glass o |Explanation / Answer
We know that mirror equation is
1/f = 1/v + 1/u
f = v*u/(v + u)
Now initially in case 1:
u1 = object distance
v1 = image distance
f = focal length of mirror, then
f = v1*u1/(v1 + u1)
Also Given that
v1/u1 = 2
v1 = 2*u1
Using above value
f = 2*u1*u1/(2*u1 + u1) = 2*u1/3
Now when both object and screen are moved, then
u2 = new object distance
v2 = new image distance
f = focal length of mirror, then
f = v2*u2/(v2 + u2)
Also Given that
v2/u2 = 3
v2 = 3*u2
Using above value
f = 3*u2*u2/(3*u2 + u2) = 3*u2/4
Since focal length of mirror will be same, So
2*u1/3 = 3*u2/4
u2 = 8*u1/9
Now also given that
v2 = v1 + 75 cm
Since v2 = 3*u2 & v1 = 2*u1 & u2 = 8*u1/9
3*u2 = 2*u1 + 75 cm
3*8*u1/9 = 2*u1 + 75
8*u1/3 - 2*u1 = 75
2*u1/3 = 75 cm
u1 = initial object distance
u1 = 75*3/2 = 112.5 cm
Now using this values
u2 = 8*112.5/9 = 100 cm
So object is moved by 112.5 cm - 100 cm = 12.5 cm
Now from above
f = 2*u1/3 = 2*112.5/3 = 75 cm
focal length = 75 cm
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