Question

6) A converging lens with focal length 15 cm, a 10 cm object locates at 30 cm from the lens, a) draw the picture and locate the image, b) calculate the image distance and the image height. c) calculate magnification, whether the image is smaller or larger, d) whether the image is inverted or upright, real or virtual. Draw picture

wave

7. What are the smallest thicknesses of a soap bubble that produce constructive interference visible light of 380 nm (violet) and 760 nm(red).? And (b) the smallest thicknesses will give destructive interference? The index of refraction of soap is 1.33. Draw picture

Answer 1

a)

lens formula  

do = 2*F = 2*15 = = -30 cm , f = 15 & ho = 10cm

( here we can see that object at 2*F so we can directly say that iamge will alos be on 2F)

1/di -1/do = 1/f

1/di -1/-30 = 1/15

di = 30 cm other side of the object ( because of + sign )  

magnification = di/do = 30/-30 = -1 negative magnification means image is inverted and real

height of image = ho* m = 10*1 = 10 cm same size  

soap bubble  

constructive interference

2*t = m *lambda/2 for minimum m = 1

t_min = lambda/ 4 = 380*10^-9/4 =9.5e-8 = 95 nm

for destructive interference

t_min = 1*lambda/ 2 = 380*10^-9/2 = 190 nm

let me know in a comment if there is any problem or doubts