A) An object is placed 51 cm to the left of a converging lens of focal length 18

Question

A) An object is placed 51 cm to the left of a converging lens of focal length 18 cm. A diverging lens of focal length 20 cm is located 18 cm to the right of the first lens. (Consider the lenses as thin lenses).
1) Where is the final image with respect to the second lens? _____cm
2) What is the linear magnification of the final image? ______

B) A converging lens f1=10cm, is placed 28cm to the left of a diverging lens, f2=15cm. An object is placed 19cm to the left of the converging lens.

Locate the final image with respect to the second lens. ________cm
and find its magnification ______ x

Explanation / Answer

A) u = object distance from lens = 0.51m

f=focal length = 0.18m

location of divergent lens = 0.18 m from convergent lens

f' = focal length of divergent lens = -0.20 mrs

v=?

1) Let us first locate image due to convex lens

v1 = image position of convex lens = uf/u-f = 51cm X 18cm/ (51+18) cm= 13.3 cm

magnification m1 = -v/u =- 13.3/ 51.0 = - 0.26 (inverted -diminished-real image)

Now this image becomes object for Diverging lens

focal length =-20cm

u ' = obect distance = 18- 13.3 = 4.7 cm (from D lens-on left side of D lens)

v2= image position of second lens = f ' u' / ( u ' - f ') = (-20 cm) X (4.7) / (4.7- (-20.0)) =-3.8 cm

Location of final image with reference to D-lens = -3.8 cm to left of D-lens

m2 = magnification = -V2/u2 = -(-3.8/4.7) = 0.81

Net magnification=(-0.26) (0.81) = - 0.21

Note : I am allowed by chegg to solve only 1 problem and it's sub-parts-sotrry.

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