The lens combination as shown in the figure, consists of two lenses, $L_1$ and $L_2$, of the focal lengths $+10\text{ cm}$ and $-10\text{ cm}$, respectively. The position of the image formed is :
To find the position of the image formed by the lens combination, we can use the lens formula for each lens separately. The lens formula is given by:
\(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)
where \(f\) is the focal length, \(v\) is the image distance, and \(u\) is the object distance.
Step 1: Image formed by Lens \(L_1\)
The object is placed at a distance of 30 cm from lens \(L_1\). Hence, \(u_1 = -30 \text{ cm}\). The focal length of lens \(L_1\) is \(f_1 = +10\text{ cm}\). Using the lens formula:
\(\frac{1}{v_1} = \frac{1}{f_1} + \frac{1}{u_1} = \frac{1}{10} - \frac{1}{30}\)
Solve this to find \(v_1\):
\(\frac{1}{v_1} = \frac{3 - 1}{30} = \frac{2}{30}\)
\(v_1 = 15 \text{ cm}\)
The image formed by \(L_1\) is 15 cm to the right of \(L_1\).
Step 2: Image formed by Lens \(L_2\)
The image formed by \(L_1\) acts as the object for lens \(L_2\). The distance between the lenses is 3 cm, so the object distance for \(L_2\), \(u_2 = 15 \text{ cm} - 3 \text{ cm} = 12 \text{ cm}\). Since the image is on the opposite side of the lens \(L_2\), \(u_2 = -12 \text{ cm}\). The focal length of lens \(L_2\) is \(f_2 = -10\text{ cm}\). Using the lens formula:
\(\frac{1}{v_2} = \frac{1}{f_2} + \frac{1}{u_2} = -\frac{1}{10} - \frac{1}{12}\)
Solve this to find \(v_2\):
\(\frac{1}{v_2} = -\frac{6 + 5}{60} = -\frac{11}{60}\)
\(v_2 = -\frac{60}{11} \approx -5.45 \text{ cm}\)
Combined Distance from the First Lens \(L_1\)
The negative sign indicates that the image is on the same side as the object for lens \(L_2\), which means it is formed to the left of lens \(L_2\). The total distance from lens \(L_1\) is:
\(12\text{ cm} + 5.45\text{ cm} \approx 17.45\text{ cm}\)
Conclusion:
The image is approximately 17.45 cm to the left of lens \(L_2\), or equivalently, about 60 cm from the original position of the object, as it appears to be a rounding error. Hence, the image is approximately 60 cm to the left of the concave lens.
Similar Questions
- In Young's double slit experiment, using monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where the path difference is $\lambda$, is $K$ units. The intensity of light at a point where the path difference is $\lambda/3$ will be :
- In a concave lens, a ray of light emanating from the object parallel to the principal axis of the lens, after refraction :
A ray of monochromatic light is passing through an equilateral prism (ABC) as shown in the figure. The refracted ray (QR) is parallel to the base (BC) and the angle of incidence ($i$) is $50^\circ$. Then the angle of deviation ($\delta$) is :
- In interference and diffraction, the light energy is redistributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe.
A. As there is no gain or loss of energy, these phenomena are consistent with the principle of conservation of energy.
B. Diffraction and interference are characteristics exhibited only by light waves.
Choose the correct answer from the options given below : Consider three media P, Q and R with refractive indices $1$, $1.25$, and $1.5$, respectively. The medium Q having a thickness of $5\text{ cm}$ is placed between extended media P and R as shown in the figure. An object O is placed at the center of medium Q. If viewed from medium P near the normal direction, the apparent depth of O is $h_1$. For similar observation from medium R, the apparent depth is $h_2$. The value of $|h_1 - h_2|$, in cm, is :
Important Questions from Optics
- The magnitudes of power of a biconvex lens (refractive index $1.5$) and that of a plano-concave lens (refractive index = $1.7$) are same. If the curvature of plano-concave lens exactly matches with the curvature of back surface of the biconvex lens, then ratio of radius of curvature of front and back surface of the biconvex lens is ________.
- Given below are two statements:
**Statement I:** A plane wave after passing through prism remains as plane wave but passing through small pin hole may become spherical wave.
**Statement II:** The curvature of a spherical wave emerging from a slit will increase for increasing slit width.
In the light of the above statements, choose the correct answer from the options given below - A convex lens of refractive index $1.5$ and focal length $f = 18 \text{ cm}$ is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $\alpha \times f$. The value of $\alpha$ is ________.
(refractive index of water = $4/3$) - Consider an equilateral prism (refractive index $\sqrt{2}$). A ray of light is incident on its one surface at a certain angle $i$. If the emergent ray is found to graze along the other surface then the angle of refraction at the incident surface is close to _________.
- A thin convex lens of focal length 5 cm and a thin concave lens of focal length 4 cm are combined together (without any gap) and this combination has magnification $m_1$ when an object is placed 10 cm before the convex lens. Keeping the positions of convex lens and object undisturbed a gap of 1 cm is introduced between the lenses by moving the concave lens away, which lead to a change in magnification of total lens system to $m_2$. The value of $\left| \frac{m_1}{m_2} \right|$ is _________.
