Physics question below?

A diverging lens has a focal length of -31.0 cm. An object is placed 18.0 cm in front of this lens.





a) Calculate the image distance.


b) Calculate the magnification.


c) Is the image real or virtual?


d) Is the image upright or inverted?


e) Is the image enlarged or reduced in size?

Physics question below?
________________________________________...


In S.I.system,focal length 'f' of diverging (concave ) lens is negative and object distance 'u' is negative





'f'= -31 cm





'u'= -18 cm





v=?





1/v - 1/u=1/f





1/v=1/u + 1/f=1/( - 18) + 1 /(-31)= 49/558





v= { - 558 }/49 = - 11.18 cm





Negative 'v' indicates image on same side as object





Magnification 'm' = v/u = [ -558/49] / 18 =0.632





Magnification is positive and less than one means image is smaller, upright,virtual.


(a)11.8 cm


(b)0.632


(c)virtual


(d) upright


(e)reduced in size


______________________________________...
Reply:1/object + 1/image = 1/focus





They give you object and focus. Solve for image.





i = 1/ (1/f - 1/o) = of / (o-f)





b) magnification = i/o (whether m is positive or negative tells you (d). Whether or not m's magnitude is greater than or less than 1 tells you (e))





c and d) Draw the ray diagram. Does light actually pass the point?