Manipal Entrance Test 2016 Physics Paper

For refraction at first surface,
u=−75‌cm,R1=25‌cm
µ1=1,µ2=

4

3

∴

µ2

v′

−

µ1

u

=

µ2−µ1

R1

⇒

4∕3

v′

−

1

(−75)

=

4∕3−1

25

⇒

4

3v′

=

1

75

−

1

75

=0
⇒v′=∞
It means the ray is parallel within the sphere.
For refraction at second surface,
u=−∞,µ1=

4

3

µ2=1,R2=−25‌cm
∴

µ2

v′

−

µ1

u

=

µ2−µ1

R2

gives

1

v′

−

4∕3

(−∞)

=

1−( 4 3 )

(−25)

⇒v=+7.5‌cm
i.e. Image is formed at distance 7.5cm to the right of P2. Hance, distance of final image P from centre C=75+25=10cm