Question

Which of the following is incorrect?
${\text{A}}{\text{.}}$ A thin convex lens of the focal length ${f_1}$ is placed in contact with a thin concave lens of focal length ${f_2}$ . The combination will act as a convex lens if ${f_{1\;}} < {\text{ }}{f_2}$ .
${\text{B}}{\text{.}}$ Light on reflection on water – glass boundary will undergo a phase change of $\pi$.
${\text{C}}{\text{.}}$ Spherical aberration is minimized by an achromatic lens.
${\text{D}}{\text{.}}$ If the image of a distant object is formed in front of the retina then the defect of vision may be myopia.

Answer 1

If a thin convex lens and a thin concave lens are in contact, and if the power of the convex lens is greater than that of the concave lens, then the combination will act as a convex lens. Now try to find the power and focal length of a lens of their relation between them.
Reflation from a medium of a higher refractive index to a lower one does ${180^0}$ phase change and for the opposite incident, no phase change occurs.

Achromatic lense is used to correct the chromatic and spherical aberration.

Myopia is a common vision condition in which we can see nearby objects clearly but far object blur

Complete step by step answer:
In this question we have to find the incorrect statement,

Now we compared from the given options and also consider the options one by one,
${\text{A}}{\text{.}}$ A thin convex lens of the focal length ${f_1}$ is placed in contact with a thin concave lens of focal length ${f_2}$ . The combination will act as a convex lens if ${f_{1\;}} < {\text{ }}{f_2}$ :
If a thin convex lens and a thin concave lens are in contact, and if the power of the convex lens is greater than that of the concave lens then the combination will act as a convex lens. Since the focal length of a lens is inversely proportional to the power of the lens, then we can say for the lesser value of the focal length of a convex lens i.e ${f_{1\;}} < {\text{ }}{f_2}$ , the combination acts as a convex lens. So the statement is correct. Hence this is not the right answer.

${\text{B}}{\text{.}}$ Light on reflection on water – glass boundary will undergo a phase change of $\pi$:
Reflation from a medium of a higher refractive index to a lower one does ${180^0}$ (=$\pi$) phase change and from the lower to higher there is no phase change. Since, the refractive index of water is less than that of the glass that means the light reflects from the medium of a lower refractive index to a higher one, so there is no phase change occurs.
Hence the statement is incorrect and this is the right option.

${\text{C}}{\text{.}}$ Spherical aberration is minimized by an achromatic lens :
The outer parts of a lens occur the spherical aberration that it can not bring light rays and also the same focus at the central part of a lens. The achromatic lens is used to limit the effects of spherical aberration. So this is the right statement. Hence this is not the right option.

${\text{D}}{\text{.}}$ If the image of a distant object is formed in front of the retina then the defect of vision may be myopia :
Myopia is a common vision condition in which we can see the nearest image or objects clearly but we can see the far object blur. So, the distant object is formed in front of the retina instead of on the retina. This is the right statement. Hence this is not the right option.

Hence the correct option is $\left( B \right)$.

Note: If we take a thin convex lens of the focal length ${f_1}$ is placed in contact with the another thin concave lens of focal length ${f_2}$, we have to check the two focal lengths with their equivalent focal length their relation between them, say $F$ ,$\dfrac{1}{F} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$
Equivalent focal length $F$ can be determined from the lens equation, $\dfrac{1}{F} = \dfrac{1}{v} - \dfrac{1}{u}$ where $v$ is the image distance from the lens and $u$ is the object distance from the lens.