Consider the actual height of the tumbler be H.

The refractive index of the water, $$\mu$$ = 4/3

The refractive index of the air, $$\mu$$ = 1

As we know that,

$${\mu _{water}} = {{{H_{real}}} \over {{H_{apparent}}}} \Rightarrow {4 \over 3} = {H \over {{H_{apparent}}}}$$

$$ \Rightarrow {H_{apparent}} = {{3H} \over 4}$$

Height of air observed by observer = 17.5 $$-$$ H

According to question, both height observed by observer is same.

$${{3H} \over 4}$$ = 17.5 $$-$$ H

$$\Rightarrow$$ H = 10 cm

Option (b)

Shortest distance is 2a between I₁ & I₃

But answer given is for I₁ & I₂

= $$\sqrt {{{(4a)}^2} + {{(2a)}^2}} $$

= $$a\sqrt {20} $$

= 4.47 a

Option (b)

$${I_{\max }} = {\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2}$$

= 4 I₀

$${1 \over {{f_1}}} = ({\mu _1} - 1)\left( {{1 \over R}} \right)$$

$${1 \over {{f_2}}} = ({\mu _2} - 1)\left( {{1 \over R}} \right)$$

$${1 \over {{f_1}}} + {1 \over {{f_2}}} = {1 \over {{f_{eq}}}} = {{({\mu _1} - 1) - ({\mu _2} - 1)} \over R}$$

$${1 \over {{f_{eq}}}} = {{({\mu _1} - {\mu _2})} \over R}$$

$${R \over {{f_{eq}}}} = ({\mu _1} - {\mu _2})$$