Two identical equiconvex thin lenses each of focal lengths 20 cm, made of material of refractive index 1.5 are placed coaxially in contact as shown. Now, the space between them is filled with a liquid with a refractive index of 1.5. The equivalent power of this arrangement will be:

2. \(R\)
3. \(\frac{3}{2}R\)
4. \(R^2\)

Shown in the figure here is a convergent lens placed inside a cell filled with a liquid. The lens has focal length + 20 cm when in air and its material has refractive index 1.50. If the liquid has refractive index 1.60, the focal length of the system is

1. + 80 cm                         

2. – 80 cm

3. – 24 cm                         

4. –160 cm

A plane-convex lens fits exactly into a plano-concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices μ₁ and μ₂ and R is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is:
1. R/2(μ_{1 +} μ₂)
2. R/2(μ_{1 -} μ₂)
3. R/(μ_{1 -} μ₂)
4. 2R/(μ₂ - μ₁)

Two identical thin plano-convex glass lenses (refractive index = \(1.5\)) each having radius of curvature of \(20\) cm are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of a refractive index of \(1.7\). The focal length of the combination is:
1. \(-20\) cm
2. \(-25\) cm
3. \(-50\) cm
4. \(50\) cm

Two similar thin equi-convex lenses, of focal length \(f\) each, are kept coaxially in contact with each other such that the focal length of the combination is \(F_1\)${\mathrm{}}_{}$. When the space between the two lenses is filled with glycerin which has the same refractive index as that of glass \((\mu = 1.5),\) then the equivalent focal length is \(F_2\)${\mathrm{}}_{}$. The ratio \(F_1:F_2\) will be:
1. \(3:4\)
2. \(2:1\)
3. \(1:2\)
4. \(2:3\)

A double convex lens has a focal length of \(25\) cm. The radius of curvature of one of the surfaces is double of the other. What would be the radii if the refractive index of the material of the lens is \(1.5?\)

The focal length of a glass $\left(\mu =1.5\right)$ lens in air is 20 cm. If it is dipped in water $\left(\mu =\frac{4}{3}\right)$, its focal length in water will be:

A lens of focal length \(f_{a}\) in air consists of a glass of refractive index \(\mu_{g}.\) If \(f_{l}\) is its focal length in a liquid of refractive index \(\mu_{l},\) then for \(\mu_{l}=\mu_{g}\)
1. \(f_{l}=0\)
2. \(\infty >f_{l}>f_{a}\)
3. \(0<f_{l}<f_{a}\)
4. \(f_{l}=\infty \)

Double-convex lenses are to be manufactured from a glass of refractive index \(1.55\) with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be \(20~\text{cm}\)?
1. \(20~\text{cm}\)
2. \(22~\text{cm}\)
3. \(24~\text{cm}\)
4. \(15~\text{cm}\)