Explanation

For a thin prism, the angle of deviation is given by

$$ \delta = (\mu - 1)A $$

For dispersion without deviation, the net deviation produced by the two prisms must be zero.

Hence, the deviation produced by the first prism must be equal and opposite to the deviation produced by the second prism.

Let the angle of the second prism be $A_2$.

Deviation due to first prism:

$$ \delta_1 = (1.72 - 1)\times 5^\circ $$

$$ \delta_1 = 0.72 \times 5^\circ = 3.6^\circ $$

Deviation due to second prism:

$$ \delta_2 = (1.9 - 1)A_2 $$

For zero net deviation,

$$ \delta_1 = \delta_2 $$

$$ 0.72 \times 5 = 0.9 \times A_2 $$

$$ A_2 = \frac{3.6}{0.9} $$

$$ A_2 = 4^\circ $$

Therefore, the angle of the second prism is

$$ \boxed{4^\circ} $$

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