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,
μYv−μXu=μY−μXR
Where, v is the image distance.
∴32v−43−x=32−43−10=−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.