Lenses - Short Question Solution | Physics Grade XI

Short Question Solution

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Lenses - Short Question Solution | Physics Grade XI

Lenses - Short Question Answers

Short Question Answers
 
1. If a converging lens and a diverging lens having the same focal length be in contact, how will the combination of lenses have?
If f1 and f2 be the focal length of converging and diverging lenses respectively, then their combined focal length is given by the relation 1/F = 1/f1 - 1/f2, where the focal length of the concave lens is taken as negative. Since f1 = f2 = f, F = 0. The combination acts as a plane glass slab.
 
2. A convex lens is immersed in water. Will its focal length change?
Yes, the focal length of the convex lens when immersed in water changes because according to lens maker formula, focal length in the air is given by
     1/fa = (aµg – 1) [(1/ r1) + (1/ r2)] …………… (i)
Where aµg is the refractive index of glass w.r.t air, fa is the focal length in air, r1 and r2 are the radii of the curvature of the two surfaces.
 
Focal length, when immersed in water, is given by
     1/fw = (wµg – 1) [(1/ r1) + (1/ r2)] ……..... (ii)
Where wµg is the refractive index of glass w.r.t water, fw is the focal length in air, r1 and r2 are the radii of curvature of the two surfaces.
 
Dividing (i) by (ii), we get
fw/fa = (aµg – 1) [(1/ r1) + (1/ r2)] / (wµg – 1) [(1/ r1) + (1/ r2)]
 
On solving and putting value of wµgaµg / wµg,
fw = 3.9fa
 
3. Sunglasses are curved but their power is 0, why?
Sunglasses have curved faces or surfaces, but their power is zero because they have two curved surfaces, one of them is convex and another is concave of the same powers and the same radii of curvatures. So, r1 = -r2 = r, then from lens maker formula
     1/f = (µ – 1) [(1/ r2) - (1/ r2)] = 0
   or, P = 0 i.e. the power of sunglasses is zero

4. Draw the ray diagram showing the formation of a real image by a concave lens. 

Real image by concave lens When the converging beam of light is incident on a concave lens, the rays of light are diverged and meet at a point in the principal axis as shown in the figure above. In this case, the object is virtual and the image of the object is real.

 

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