Respuesta :
-0.233 m left of the diverging lens and ( 0.12 - 0.233 ) = -113 m left of the converging lens and 0.023 m right of the diverging lens.
What is lens formula?
The lens formula is
[tex]\dfrac{1}{f} =\dfrac{1}{u} + \dfrac{1}{v}[/tex]
Here, f is the focal length, u is the distance of the object and v is the distance of the image.
The given data is follows as
focal length f2 = 14 cm = -0.14 m
Separation s = 12 cm = 0.12 m
focal length f1 = 21 cm = 0.21 m
The distance u1 = 38 cm
we need to find the image location i.e v2
so by lens formula v1 is
[tex]\dfrac{1}{f} =\dfrac{1}{u} + \dfrac{1}{v} v_1 = 1/(1/f_1 - 1/u_1)\\\\v_1 = 1/( 1/0.21 - 1/0.38)\\\\v_1 = 0.47 m[/tex]
[tex]u_2 = s - v_1\\u_2 = 0.12 - 0.47\\u_2 = -0.35[/tex]
so from equation 1
[tex]v_2 = 1/(1/f_2 - 1/u_2)\\\\v_2 = 1/(-1/0.14 + 1/0.35)\\\\v_2 = -0.233 m[/tex]
So, -0.233 m left of diverging lens and ( 0.12 - 0.233 ) = -113 m left of converging and for Separation s = 45 cm = 0.45 m
[tex]v_1 = 1/(1/f_1 - 1/u_1)\\v_1 =0.47 m\\and\\u_2 = s - v_1\\u_2 = 0.45 - 0.47 =- 0.02 m\\so\\v_2 = 1/(1/f_2 - 1/u_2)\\v_2 = 1/(-1/0.14 + 1/0.02)\\v_2 = 0.023[/tex]
so here is 0.023 m right of a diverging lens.
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