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Question

A concave spherical surface of radius of curvature 10cm separates two mediums X and Y of refractive indices 4/3 and 3/2 respectively. Centre of curvature of the surface lies in the medium X. An object is placed in medium X.

A
Image is always real
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B
Image is real if the object distance is greater than 90cm
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C
Image is always virtual
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D
Image is real if the object distance is less than 90cm
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Solution

The correct option is C Image is always virtual
It is given, that, a concave spherical surface of radius of curvature 10cm separates two medium of X and Y of refractive index 43 and 32 respectively.

Therefore, μX=43 , μY=32

And, according to sign convention, R=10cm.

Now, let us consider, the object is situated at a distance of x in front of the concave spherical surface. So, object distance, u=x (using sign convention)

According to lens make's formula, we know,

μYvμXu=μYμXR

Where, v is the image distance.

32v43x=324310=160
Or, v=1.5x0.017x+1.33

So, clearly, v<0

This implies that the image will always be a virtual one.

Therefore, the correct option is - (C) The image is always virtual.

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