An equiconvex lens (of refractive index 1.5) is placed in contact with plane mirror as shown in the fogure. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of needle. the distance of the needle from the lens is measured to be 30 cm. Now a few drops of a liquid are put in between the lens and the plane mirror and the new position of the images is found to be located at 45 cm. calculate the refractive index of the liquid.
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Given,
focal length, f = -15 cm
Distance of image, v = -10 cm
Distance of object, u =?
We know that,
1/v - 1/u = 1/f
So,
1/-10 - 1/u = 1/-15
So, -1/10 + 1/15 = 1/u
(-3 + 2)/30 = 1/u
-1/30 = 1/u
Therefore u = -30 cm
Thus object is placed 30 cm away from the concave lens.
Negative sign shows that object is at 30cm in front of the lens.
focal length, f = -15 cm
Distance of image, v = -10 cm
Distance of object, u =?
We know that,
1/v - 1/u = 1/f
So,
1/-10 - 1/u = 1/-15
So, -1/10 + 1/15 = 1/u
(-3 + 2)/30 = 1/u
-1/30 = 1/u
Therefore u = -30 cm
Thus object is placed 30 cm away from the concave lens.
Negative sign shows that object is at 30cm in front of the lens.
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