Given that, d = 1.5m; D = 25cm = 0.25m (For a normal eye).
From equation (2.8), the focal length of the correction lens is
\(f=\cfrac { d\times D }{ d-D } =\cfrac { 1.5\times 0.25 }{ 1.5-0.25 } =\cfrac { 0.375 }{ 1.25 } =0.3m\)

According to Snell’s law,
\(\frac{\sin i}{\sin r}=\frac{\mu_2}{\mu_1}\)
\(\mu_1 \sin i=\mu_2 \sin r\)
Here \({ \mu }_{ 1 }=1.0,{ \mu }_{ 2 }=1.5,i\ ={ 30 }^{ 0 }\)
\(\begin{aligned} & (1.0) \sin 30^0=1.5 \sin r \end{aligned}\)
\(\begin{aligned} 1 \times \frac{1}{2}=1.5 \sin r \end{aligned}\)
\(\sin r=\frac{1}{2 \times 1.5}=\frac{1}{3}=(0.333)\)
r = sin^-1(0.333)
r = 19.45^o

Given:
\(\mathrm{f}=10 \mathrm{~cm}, \mathrm{u}=-20 \mathrm{~cm}, \mathrm{v}=?  \)
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}  \Rightarrow \frac{1}{v} =\frac{1}{f}+\frac{1}{u} \)
\(\frac{1}{v} =\frac{1}{10}+\frac{1}{-20}=\frac{1}{10}-\frac{1}{20}  \)
\(\frac{1}{v} =\frac{2-1}{20}=\frac{1}{20}  \)
\(\mathbf{v} =20 \mathrm{~cm}  \)
Image distance = 20 cm
Nature of image:
Nature of the image is real, enlarged and inverted image.

(i) Light is a form of energy.
(ii) Light always travels along a straight line.
(iii) Light does not need any medium for its propagation. It can even travel through vacuum.
(iv) The speed of light in vacuum or air is, c = 3 x 10⁸ m/s
(v) Since, light is in the form of waves, it is characterized by a wavelength (λ) and a frequency (v), which are related by the following equation: c = v λ (c - velocity of light).