I hv 5 questions answer all if u can or im satisified with 2 or 3.Here are my questions:
1>A convex lens of focal length 16cm forms a real image double the size of the object.The distance of the object from the length is?
2.A convex lens produces a real image m times size of the object.The distance of the object from the lens is?
3>A 2.0cm tall object is placed perpendicular to prinicipal axis of convex lens of focal length 10cm, then nature of image is?
4>A convex lens is placed somewhere in btwn an object and screen.The distance btwn the object and screen is 48cm.if the numerical value of magnification produced by lens is 3, focal length of lens is?
5>The diameter of sun is 1.4×10^9 and its distance frm earth is 1.5×10^11m.Radius of image of sun formed by a lens of focal length is?
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Ans1 f= -16cm
, m=-v/-u=2
; v=2u :
1/f = 1/v - 1/u ; ;
1/-16 = 1/-2u - 1/-u ;
-1/16 = (-1+2)/2u ;
-16 =2u ;
u = -8cm
Ans 2 Given,
-u + v = 100 ......eq (1)
Magnification = v /u = -3 (real and inverted image)
v = -3u ........eq (2)
Focal length 1/f = 1/v - 1/u ......eq (3)
Substitute eq (2) in eq (1), we get u = -25 cm
eq (2) v = -3 u = - 3 *- 25 = 75 cm
Substitute eq (2) in eq (3), we get f = - 3 u/4 = -3 * -25/4 = 18.75 cm.
, m=-v/-u=2
; v=2u :
1/f = 1/v - 1/u ; ;
1/-16 = 1/-2u - 1/-u ;
-1/16 = (-1+2)/2u ;
-16 =2u ;
u = -8cm
Ans 2 Given,
-u + v = 100 ......eq (1)
Magnification = v /u = -3 (real and inverted image)
v = -3u ........eq (2)
Focal length 1/f = 1/v - 1/u ......eq (3)
Substitute eq (2) in eq (1), we get u = -25 cm
eq (2) v = -3 u = - 3 *- 25 = 75 cm
Substitute eq (2) in eq (3), we get f = - 3 u/4 = -3 * -25/4 = 18.75 cm.
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