JEE MAIN - Physics (2016 - 10th April Morning Slot - No. 22)
A hemispherical glass body of radius 10 cm and refractive index 1.5 is silvered on its curved surface. A small air bubble is 6 cm below the flat surface inside it along the axis. The position of the image of the air bubble made by the mirror is seen :
_10th_April_Morning_Slot_en_22_1.png)
14 cm below flat surface
30 cm below flat surface
20 cm below flat surface
16 cm below flat surface
คำอธิบาย
_10th_April_Morning_Slot_en_22_2.png)
Using mirror formula.
$${1 \over v} + {1 \over { - 4}} = {1 \over { - 5}}$$
$$ \Rightarrow \,\,\,{1 \over v} = {1 \over 4} - {1 \over 5} = {1 \over {20}}$$
$$ \Rightarrow \,\,\,v = 20\,$$ cm
$$ \therefore $$ Image distance is 20 cm
_10th_April_Morning_Slot_en_22_3.png)
I1 acts as object for plane glass surface.
$$ \therefore $$ Appartent depth = $${{R + v} \over \mu } = {{30} \over {1.5}} = 20$$ cm
$$ \therefore $$ Position of the image of the air bubble is 20 cm below the flat surface.