19.. Which pair of most accurately describes the image? A) terms B) real, C) rea

Question

19.. Which pair of most accurately describes the image? A) terms B) real, C) real, inverted D) virtual, inverted E) Ans: E. Refer Ref 26-8 Difficulty: Medium Section Def: Section 26-6, 26-7,26-8 Reference: Ref 26-9 A 1.5-cm object is placed 0.50 m to the left of a diverging lens with a length of 020 m. A converging lens with a focal length of 0.17 m is located 0.08 m to the right of the diverging lens. 20.. What is the location of the final image with respect to the object? A) The final image is located 0.14 m to the left of the object. B) The final image is located 0.32 m to the right of the object. C) The final image is located 0.40 m to the right of the object. D) The final image is located 0.83 m to the right object. E) The final image is located 1.3 m to the right of the object. Ans: E Refer To: Ref 26-10 Difficulty: Hard SectionDef Section 26-9 21. What is the height and orientation with respect to the original object of the final image? A) 1.4 cm, inverted B) 1.4 cm, upright C) 0.95 cm, inverted D) 0.95 cm, upright E) 0.28 cm, inverted Ans: A

Explanation / Answer

19)
option A) false

option B) false

option C) true

option D) false

option E) true

here it seem to be both (C) and (E) are correct, but most accuracy is option (E)

20)

caluculation of image due to first lense:

object distance u=0.5m

f1=-0.2m

let image distance be v

1/u+1/v=1/f1

1/0.5+1/v=1/-0.2

===>

v=-0.14 m this is the image due to first lense

now caluculation of image due to second lense:

object distance for the second lense is,

u'=0.08+0.14

u'=0.22 m

let image distnace be v'

and

f2=0.17 m

now,

1/u' +1/v'=1/f2

1/0.22+1/v'=1/0.17

===>

v'=0.74 m

image distance due to second lense is v'=0.74 m

now,

final image distance from the actual object is=0.5+0.08+0.+0.74=1.32 m

hence option (E) is correct

21)

total magnification M=m1*m2=h'/h

here,

object height h=1.5 cm

final image height is h'

and

magnificatin of first lense is m1=v/u=-0.14/0.5=-0.28

magnificatin of second lense is m2=v'/u'=0.74/0.22=3.36

now,

m1*m2=h'/h

===>

h'=h*(m1/m2)

=1.5*(-0.28/3.36)

h'=-1.14 cm

hence image is inverted and correct option is (A)

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