A convex lens focuses an object 20 cm from i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The apparent shift in the position of the image due to the glass slab is given by $d = t(1 - \frac{1}{\mu})$.Substituting the values, $d = 1.4(1 -

Vedclass · Accepted Answer

$30.7$

Vedclass · Answer

(B) The apparent shift in the position of the image due to the glass slab is given by $d = t(1 - \frac{1}{\mu})$.Substituting the values, $d = 1.4(1 - \frac{5}{7}) = 1.4(\frac{2}{7}) = 0.4 \ cm$.First, calculate the focal length $f$ of the lens using the initial conditions: $u = -20 \ cm$ and $v = 5 \ cm$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$, we get $\frac{1}{f} = \frac{1}{5} - \frac{1}{-20} = \frac{4+1}{20} = \frac{5}{20} = \frac{1}{4}$. Thus, $f = 4 \ cm$.When the glass slab is inserted, the image must still form on the screen at $v = 5 \ cm$. However, the slab shifts the image position towards the lens by $d = 0.4 \ cm$. Therefore, the effective image distance becomes $v' = 5 - 0.4 = 4.6 \ cm$.Using the lens formula again for the new object distance $u'$: $\frac{1}{f} = \frac{1}{v'} - \frac{1}{u'}$.$\frac{1}{4} = \frac{1}{4.6} - \frac{1}{u'} \Rightarrow \frac{1}{u'} = \frac{1}{4.6} - \frac{1}{4} = \frac{4 - 4.6}{18.4} = \frac{-0.6}{18.4}$.$u' = -\frac{18.4}{0.6} \approx -30.66 \ cm \approx 30.7 \ cm$.

Similar Questions

In the figure,an air lens of radii of curvature $10 \ cm$ ($R_1 = 10 \ cm$,$R_2 = -10 \ cm$) is cut in a cylinder of glass $(\mu = 1.5)$. The focal length and the nature of the lens is:

$A$ light ray hits the pole of a thin biconvex lens as shown in the figure. The angle made by the emergent ray with the optic axis will be

$A$ convex lens is made from glass material having a refractive index of $1.4$ with the same radius of curvature on both sides. The ratio of its focal length to its radius of curvature is . . . . . . .

If in a plano-convex lens,the radius of curvature of the convex surface is $10 \,cm$ and the focal length of the lens is $30 \,cm$,the refractive index of the material of the lens will be ..........

$Assertion (A)$: The focal length of a lens does not change when red light is replaced by blue light.
$Reason (R)$: The focal length of a lens does not depend on the colour of the light used.

Vedclass Test Series

Exam Paper Generator

Online Exam Module