A plano convex lens off \(=20 \mathrm{~cm}\) is silvered at plane surface New f will be \(\quad \mathrm{cm}\) (A) 20 (B) 40 (C) 30 (D) 60

Short Answer

Expert verified
The new focal length of the plano-convex lens after silvering is \(40\) cm (option B). This is because the silvered surface causes internal reflection, effectively making the light pass through the lens twice and forming a virtual image at twice the original focal length.

Step by step solution

01

Understand the effect of silvering on plano-convex lens

When the plane surface of a plano-convex lens is silvered, the light entering the lens will be internally reflected by the silvered surface. This internal reflection causes the light to effectively pass through the lens twice, forming a virtual image at twice the original focal length.
02

Apply the lens formula

The lens formula is given by: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\) , where \(f\) is the focal length of the lens, \(v\) is the distance of the image from the lens, and \(u\) is the distance of the object from the lens. Since the image and object are at the same distance (due to internal reflection), we can replace \(v\) and \(u\) with \(x\), and the equation becomes: \(\frac{1}{f} = \frac{1}{x} - \frac{1}{x}\)
03

Obtain the new focal length

Given that the original focal length of the lens is \(20\) cm, the light effectively passes through the lens twice, doubling the distance between the lens and the virtual image. Therefore, the new focal length, denoted by \(f'\), can be obtained by multiplying the original focal length by 2: \(f' = 2f = 2 (20 \mathrm{~cm}) = 40 \mathrm{~cm}\)
04

Choose the correct answer

The new focal length of the plano-convex lens after silvering is \(40\) cm. So the correct option is (B) 40 cm.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In young's double slit experiment the phase difference is constant between two sources is \((\pi / 2)\). The intensity at a point equidistant from the slits in terms of max. intensity \(\mathrm{I}_{0}\) is (A) \(3 \mathrm{I}_{0}\) (B) \(\left(\mathrm{I}_{0} / 2\right)\) (C) I_{0 } (D) \(\left(3 \mathrm{I}_{0} / 4\right)\)

In young's double Slit experiment the seventh maxima with wavelength \(\lambda_{1}\), is at a distance \(\mathrm{d}_{1}\) and the same maxima with wavelength \(\lambda_{2}\), is at a distance \(\mathrm{d}_{2}\) Then \(\left(\mathrm{d}_{1} / \mathrm{d}_{2}\right)=\) (A) \(\left(\lambda_{2} / \lambda_{1}\right)\) (B) \(\left(\lambda_{1}^{2} / \lambda_{2}^{2}\right)\) (C) \(\left(\lambda_{2}^{2} / \lambda_{1}^{2}\right)\) (D) \(\left(\lambda_{1} / \lambda_{2}\right)\)

For a prism of refractive index \(\sqrt{3}\), the angle of minimum deviation is equitation is equal to the angle of prism, then angle of the prism is (A) \(60^{\circ}\) (B) \(90^{\circ}\) (C) \(45^{\circ}\) (D) \(180^{\circ}\)

$$ \begin{array}{|l|l|} \hline \text { Column - I } & \text { Column - II } \\ \hline \text { (i) While going from rarer to denser medium } & \text { (a) Wavelength changes } \\ \text { (ii) While going from denser to rarer medium } & \text { (b) } \eta=(\mathrm{C} / \mathrm{V}) \\ \text { (iii) While going to one medium to another } & \text { (C) Ray bends towards normal } \\ \text { (iv) Refractive index of medium } & \text { (D) Rav bends awav from normal } \\ \hline \end{array} $$ (A) \(i-c\), ii \(-d\), iii \(-b\), iv-a (B) \(\mathrm{i}-\mathrm{a}\), ii \(-\mathrm{b}\), iii $-\mathrm{c}, \mathrm{iv}-\mathrm{d}$ (C) $\mathrm{i}-\mathrm{c}, \mathrm{ii}-\mathrm{b}, \mathrm{iii}-\mathrm{a}, \mathrm{iv}-\mathrm{d}$ (D) \(i-d, 1 i-c, 11 i-b, i v-a\)

The focal lengths of objective and the eye-piece of a compound microscope are \(F_{0}\) and \(F_{e}\) respectively. Then (A) \(\mathrm{F}_{0}>\mathrm{F}_{\mathrm{e}}\) (B) \(\mathrm{F}_{0}<\mathrm{F}_{\mathrm{e}}\) (C) \(\mathrm{F}_{0}=\mathrm{F}_{\mathrm{e}}\) (D) none of these

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free