An equiconvex lens is placed in contact with plane mirror as shown in the. A small needle with its tip on 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 30cm. Now a few drops of liquid are put in between the lens and the plane mirror and the new position of the image is found to be located at 45cm. Calculate the refractive index of the liquid.
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Focal length of convex lens F1 =45
The liquid act as a mirror,focal length the liquid if F2.
Focal length of system{convex lens+liquid}F =30cm
For a pair of opticle system placed on contact.
1/f=1/f1 + 1/f2
F2=F multiply F1/ F1-f
=30multiply 45/45-30=90cm.
Let the refractive index of lens be n1.
And the radius of curvature of one surface be R, hence the radius of curvature of outer suface is -R multiply R.
1/f ={n1-1} {1/R -1/-R}
R=2F1{n-1} =2multiply 45 {1.5-1}=134cm.
Let n2 be the refractive index in liquid.
Radius of curvature of liquid on the side of lens=-R=-134cm.
The value of n2 can be using the relation 1/f2={n2-1} {1.5/-R}
n2=1 -R/f2
= 1-134/-90.
=1.488.
The liquid act as a mirror,focal length the liquid if F2.
Focal length of system{convex lens+liquid}F =30cm
For a pair of opticle system placed on contact.
1/f=1/f1 + 1/f2
F2=F multiply F1/ F1-f
=30multiply 45/45-30=90cm.
Let the refractive index of lens be n1.
And the radius of curvature of one surface be R, hence the radius of curvature of outer suface is -R multiply R.
1/f ={n1-1} {1/R -1/-R}
R=2F1{n-1} =2multiply 45 {1.5-1}=134cm.
Let n2 be the refractive index in liquid.
Radius of curvature of liquid on the side of lens=-R=-134cm.
The value of n2 can be using the relation 1/f2={n2-1} {1.5/-R}
n2=1 -R/f2
= 1-134/-90.
=1.488.
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