Optics 3 Question 2

2. A thin convex lens L (refractive index =1.5) is placed on a plane mirror M. When a pin is placed at A, such that OA=18cm, its real inverted image is formed at A itself, as shown in figure. When a liquid of refractive index μl is put between the lens and the mirror, the pin has

to be moved to A, such that OA=27cm, to get its inverted real image at A itself. The value of μl will be

(a) 3

(b) 2

(c) 43

(d) 32

(2019 Main, 9 April II)

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Answer:

Correct Answer: 2. (c)

Solution:

  1. Light from plane mirror is reflected back on it’s path, so that image of A coincides with A itself.

This would happen when rays refracted by the convex lens falls normally on the plane mirror, i.e.

the refracted rays form a beam parallel to principal axis of the lens. Hence, the object would then be considered at the focus of convex lens.

Focal length of curvature of convex lens is, f1=18cm

With liquid between lens and mirror, image is again coincides with object, so the second measurement is focal length of combination of liquid lens and convex lens.

1f1+1f2=1feq 118+1f2=127f2=54cm

For convex lens by lens maker’s formula, we have

1f=(μ1)2R118=0.5×2RR=18cm

and for plano-convex liquid lens, we have

1f=(μl1)1R154=(μl1)118μl=1+13=43



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