When light is refracted into a denser medium,
(A) its wavelength and frequency both increase
(B) its wavelength increases but the frequency remain unchanged
(C) its wavelength decreases but the frequency remain unchanged
(D) its wavelength and frequency both decrease

Hint: When the light moves from the rarer medium to the denser medium, the refractive index will be greater than 1. From this, a new relation is obtained. The frequency of the wave depends upon its nature not medium but wavelength depends upon the medium in which the wave propagates.

Formula used: The formula of the velocity of the light is given by
$v = \lambda f$
Where $v$ is the velocity of the light, $\lambda $ is the wavelength of the light and $f$ is the frequency of the light.

Complete answer: It is known that the velocity of the light decreases when the light moves from the rarer medium to the denser medium. The rarer medium is the medium that is made up of the less dense medium. The rarer medium is the medium that is made up of the denser medium.
$\mu = \dfrac{{{v_{rarer}}}}{{{v_{denser}}}} > 1$
${v_{{\text{rarer}}}} > {v_{{\text{denser}}}}$
Substituting the formula of the velocity in the above step, we get
${\lambda _1}{f_1} > {\lambda _2}{f_2}$ ………………………….(1)
It is known that the frequency of the wave depends on the nature of the wave and it is independent of the moving medium. For both the denser and the rarer medium, the value of the frequency remains the same.
So $f = {f_1} = {f_2}$
Substitute the above value in the equation (1)
$
  {\lambda _1}f > {\lambda _2}f \\
  {\lambda _1} > {\lambda _2} \\
 $
The above step shows that when the light moves from the rarer medium to the denser medium, the wavelength decreases.

Thus the option (C) is correct.

Note: Frequency is the occurrence of the wave again and again and it depends on the material in which it is made of. If the frequency of the wave is greater than the velocity of it will also be greater. The waves with different frequencies will have the same frequency.