Formation of real image using a biconvex lens is shown below : If the whole set up is immersed in water without disturbing the object and the screen positions , what will one observe on the screen ?
Hello! The answer to this is that the image will disappear . We know that
1/f=(u2/u1 - 1)(1/R1 - 1/R2)
where u1= refractive index of medium of object
u2= refractive index of lens
and R1 and R2 are the radius of curvature of the two surfaces of lens.
We know that:
1/f= 1/v - 1/u and also
1/f= (Ureal - 1)(1/R1 - 1/R2)
where Ureal= refractive index in which the set up is kept.
Now, when we immerse the whole apparatus in water then Ureal changes and as you changes f also changes. Now, the value of u will be same but f changes so the value of v also changes but the SCREEN IS KEPT AT THE SAME DISTANCE AS BEFORE so the image will not coincide on the screen and it will disappear.
Hope the answer helps.
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