Treatise on Light Part 4

You’re reading novel Treatise on Light Part 4 online at LightNovelFree.com. Please use the follow button to get notification about the latest chapter next time when you visit LightNovelFree.com. Use F11 button to read novel in full-screen(PC only). Drop by anytime you want to read free – fast – latest novel. It’s great if you could leave a comment, share your opinion about the new chapters, new novel with others on the internet. We’ll do our best to bring you the finest, latest novel everyday. Enjoy!

[Ill.u.s.tration]

28. Now pa.s.sing to the investigation of the refractions which obliquely incident rays must undergo, according to our hypothesis of spheroidal waves, I saw that these refractions depended on the ratio between the velocity of movement of the light outside the crystal in the ether, and that within the crystal. For supposing, for example, this proportion to be such that while the light in the crystal forms the spheroid GSP, as I have just said, it forms outside a sphere the semi-diameter of which is equal to the line N which will be determined hereafter, the following is the way of finding the refraction of the incident rays. Let there be such a ray RC falling upon the surface CK. Make CO perpendicular to RC, and across the angle KCO adjust OK, equal to N and perpendicular to CO; then draw KI, which touches the Ellipse GSP, and from the point of contact I join IC, which will be the required refraction of the ray RC. The demonstration of this is, it will be seen, entirely similar to that of which we made use in explaining ordinary refraction. For the refraction of the ray RC is nothing else than the progression of the portion C of the wave CO, continued in the crystal. Now the portions H of this wave, during the time that O came to K, will have arrived at the surface CK along the straight lines H_x_, and will moreover have produced in the crystal around the centres _x_ some hemi-spheroidal partial waves similar to the hemi-spheroidal GSP_g_, and similarly disposed, and of which the major and minor diameters will bear the same proportions to the lines _xv_ (the continuations of the lines H_x_ up to KB parallel to CO) that the diameters of the spheroid GSP_g_ bear to the line CB, or N.

And it is quite easy to see that the common tangent of all these spheroids, which are here represented by Ellipses, will be the straight line IK, which consequently will be the propagation of the wave CO; and the point I will be that of the point C, conformably with that which has been demonstrated in ordinary refraction.

Now as to finding the point of contact I, it is known that one must find CD a third proportional to the lines CK, CG, and draw DI parallel to CM, previously determined, which is the conjugate diameter to CG; for then, by drawing KI it touches the Ellipse at I.

29. Now as we have found CI the refraction of the ray RC, similarly one will find C_i_ the refraction of the ray _r_C, which comes from the opposite side, by making C_o_ perpendicular to _r_C and following out the rest of the construction as before. Whence one sees that if the ray _r_C is inclined equally with RC, the line C_d_ will necessarily be equal to CD, because C_k_ is equal to CK, and C_g_ to CG. And in consequence I_i_ will be cut at E into equal parts by the line CM, to which DI and _di_ are parallel. And because CM is the conjugate diameter to CG, it follows that _i_I will be parallel to _g_G. Therefore if one prolongs the refracted rays CI, C_i_, until they meet the tangent ML at T and _t_, the distances MT, M_t_, will also be equal. And so, by our hypothesis, we explain perfectly the phenomenon mentioned above; to wit, that when there are two rays equally inclined, but coming from opposite sides, as here the rays RC, _rc_, their refractions diverge equally from the line followed by the refraction of the ray perpendicular to the surface, by considering these divergences in the direction parallel to the surface of the crystal.



30. To find the length of the line N, in proportion to CP, CS, CG, it must be determined by observations of the irregular refraction which occurs in this section of the crystal; and I find thus that the ratio of N to GC is just a little less than 8 to 5. And having regard to some other observations and phenomena of which I shall speak afterwards, I put N at 156,962 parts, of which the semi-diameter CG is found to contain 98,779, making this ratio 8 to 5-1/29. Now this proportion, which there is between the line N and CG, may be called the Proportion of the Refraction; similarly as in gla.s.s that of 3 to 2, as will be manifest when I shall have explained a short process in the preceding way to find the irregular refractions.

31. Supposing then, in the next figure, as previously, the surface of the crystal _g_G, the Ellipse GP_g_, and the line N; and CM the refraction of the perpendicular ray FC, from which it diverges by 6 degrees 40 minutes. Now let there be some other ray RC, the refraction of which must be found.

About the centre C, with semi-diameter CG, let the circ.u.mference _g_RG be described, cutting the ray RC at R; and let RV be the perpendicular on CG. Then as the line N is to CG let CV be to CD, and let DI be drawn parallel to CM, cutting the Ellipse _g_MG at I; then joining CI, this will be the required refraction of the ray RC. Which is demonstrated thus.

[Ill.u.s.tration]

Let CO be perpendicular to CR, and across the angle OCG let OK be adjusted, equal to N and perpendicular to CO, and let there be drawn the straight line KI, which if it is demonstrated to be a tangent to the Ellipse at I, it will be evident by the things heretofore explained that CI is the refraction of the ray RC. Now since the angle RCO is a right angle, it is easy to see that the right-angled triangles RCV, KCO, are similar. As then, CK is to KO, so also is RC to CV. But KO is equal to N, and RC to CG: then as CK is to N so will CG be to CV. But as N is to CG, so, by construction, is CV to CD. Then as CK is to CG so is CG to CD. And because DI is parallel to CM, the conjugate diameter to CG, it follows that KI touches the Ellipse at I; which remained to be shown.

32. One sees then that as there is in the refraction of ordinary media a certain constant proportion between the sines of the angles which the incident ray and the refracted ray make with the perpendicular, so here there is such a proportion between CV and CD or IE; that is to say between the Sine of the angle which the incident ray makes with the perpendicular, and the horizontal intercept, in the Ellipse, between the refraction of this ray and the diameter CM. For the ratio of CV to CD is, as has been said, the same as that of N to the semi-diameter CG.

33. I will add here, before pa.s.sing away, that in comparing together the regular and irregular refraction of this crystal, there is this remarkable fact, that if ABPS be the spheroid by which light spreads in the Crystal in a certain s.p.a.ce of time (which spreading, as has been said, serves for the irregular refraction), then the inscribed sphere BVST is the extension in the same s.p.a.ce of time of the light which serves for the regular refraction.

[Ill.u.s.tration]

For we have stated before this, that the line N being the radius of a spherical wave of light in air, while in the crystal it spread through the spheroid ABPS, the ratio of N to CS will be 156,962 to 93,410. But it has also been stated that the proportion of the regular refraction was 5 to 3; that is to say, that N being the radius of a spherical wave of light in air, its extension in the crystal would, in the same s.p.a.ce of time, form a sphere the radius of which would be to N as 3 to 5. Now 156,962 is to 93,410 as 5 to 3 less 1/41. So that it is sufficiently nearly, and may be exactly, the sphere BVST, which the light describes for the regular refraction in the crystal, while it describes the spheroid BPSA for the irregular refraction, and while it describes the sphere of radius N in air outside the crystal.

Although then there are, according to what we have supposed, two different propagations of light within the crystal, it appears that it is only in directions perpendicular to the axis BS of the spheroid that one of these propagations occurs more rapidly than the other; but that they have an equal velocity in the other direction, namely, in that parallel to the same axis BS, which is also the axis of the obtuse angle of the crystal.

[Ill.u.s.tration]

34. The proportion of the refraction being what we have just seen, I will now show that there necessarily follows thence that notable property of the ray which falling obliquely on the surface of the crystal enters it without suffering refraction. For supposing the same things as before, and that the ray makes with the same surface _g_G the angle RCG of 73 degrees 20 minutes, inclining to the same side as the crystal (of which ray mention has been made above); if one investigates, by the process above explained, the refraction CI, one will find that it makes exactly a straight line with RC, and that thus this ray is not deviated at all, conformably with experiment. This is proved as follows by calculation.

CG or CR being, as precedently, 98,779; CM being 100,000; and the angle RCV 73 degrees 20 minutes, CV will be 28,330. But because CI is the refraction of the ray RC, the proportion of CV to CD is 156,962 to 98,779, namely, that of N to CG; then CD is 17,828.

Now the rectangle _g_DC is to the square of DI as the square of CG is to the square of CM; hence DI or CE will be 98,353. But as CE is to EI, so will CM be to MT, which will then be 18,127. And being added to ML, which is 11,609 (namely the sine of the angle LCM, which is 6 degrees 40 minutes, taking CM 100,000 as radius) we get LT 27,936; and this is to LC 99,324 as CV to VR, that is to say, as 29,938, the tangent of the complement of the angle RCV, which is 73 degrees 20 minutes, is to the radius of the Tables. Whence it appears that RCIT is a straight line; which was to be proved.

35. Further it will be seen that the ray CI in emerging through the opposite surface of the crystal, ought to pa.s.s out quite straight, according to the following demonstration, which proves that the reciprocal relation of refraction obtains in this crystal the same as in other transparent bodies; that is to say, that if a ray RC in meeting the surface of the crystal CG is refracted as CI, the ray CI emerging through the opposite parallel surface of the crystal, which I suppose to be IB, will have its refraction IA parallel to the ray RC.

[Ill.u.s.tration]

Let the same things be supposed as before; that is to say, let CO, perpendicular to CR, represent a portion of a wave the continuation of which in the crystal is IK, so that the piece C will be continued on along the straight line CI, while O comes to K. Now if one takes a second period of time equal to the first, the piece K of the wave IK will, in this second period, have advanced along the straight line KB, equal and parallel to CI, because every piece of the wave CO, on arriving at the surface CK, ought to go on in the crystal the same as the piece C; and in this same time there will be formed in the air from the point I a partial spherical wave having a semi-diameter IA equal to KO, since KO has been traversed in an equal time. Similarly, if one considers some other point of the wave IK, such as _h_, it will go along _hm_, parallel to CI, to meet the surface IB, while the point K traverses K_l_ equal to _hm_; and while this accomplishes the remainder _l_B, there will start from the point _m_ a partial wave the semi-diameter of which, _mn_, will have the same ratio to _l_B as IA to KB. Whence it is evident that this wave of semi-diameter _mn_, and the other of semi-diameter IA will have the same tangent BA. And similarly for all the partial spherical waves which will be formed outside the crystal by the impact of all the points of the wave IK against the surface of the Ether IB. It is then precisely the tangent BA which will be the continuation of the wave IK, outside the crystal, when the piece K has reached B. And in consequence IA, which is perpendicular to BA, will be the refraction of the ray CI on emerging from the crystal. Now it is clear that IA is parallel to the incident ray RC, since IB is equal to CK, and IA equal to KO, and the angles A and O are right angles.

It is seen then that, according to our hypothesis, the reciprocal relation of refraction holds good in this crystal as well as in ordinary transparent bodies; as is thus in fact found by observation.

36. I pa.s.s now to the consideration of other sections of the crystal, and of the refractions there produced, on which, as will be seen, some other very remarkable phenomena depend.

Let ABH be a parallelepiped of crystal, and let the top surface AEHF be a perfect rhombus, the obtuse angles of which are equally divided by the straight line EF, and the acute angles by the straight line AH perpendicular to FE.

The section which we have hitherto considered is that which pa.s.ses through the lines EF, EB, and which at the same time cuts the plane AEHF at right angles. Refractions in this section have this in common with the refractions in ordinary media that the plane which is drawn through the incident ray and which also intersects the surface of the crystal at right angles, is that in which the refracted ray also is found. But the refractions which appertain to every other section of this crystal have this strange property that the refracted ray always quits the plane of the incident ray perpendicular to the surface, and turns away towards the side of the slope of the crystal. For which fact we shall show the reason, in the first place, for the section through AH; and we shall show at the same time how one can determine the refraction, according to our hypothesis. Let there be, then, in the plane which pa.s.ses through AH, and which is perpendicular to the plane AFHE, the incident ray RC; it is required to find its refraction in the crystal.

[Ill.u.s.tration]

37. About the centre C, which I suppose to be in the intersection of AH and FE, let there be imagined a hemi-spheroid QG_qg_M, such as the light would form in spreading in the crystal, and let its section by the plane AEHF form the Ellipse QG_qg_, the major diameter of which Q_q_, which is in the line AH, will necessarily be one of the major diameters of the spheroid; because the axis of the spheroid being in the plane through FEB, to which QC is perpendicular, it follows that QC is also perpendicular to the axis of the spheroid, and consequently QC_q_ one of its major diameters. But the minor diameter of this Ellipse, G_g_, will bear to Q_q_ the proportion which has been defined previously, Article 27, between CG and the major semi-diameter of the spheroid, CP, namely, that of 98,779 to 105,032.

Let the line N be the length of the travel of light in air during the time in which, within the crystal, it makes, from the centre C, the spheroid QC_qg_M. Then having drawn CO perpendicular to the ray CR and situate in the plane through CR and AH, let there be adjusted, across the angle ACO, the straight line OK equal to N and perpendicular to CO, and let it meet the straight line AH at K. Supposing consequently that CL is perpendicular to the surface of the crystal AEHF, and that CM is the refraction of the ray which falls perpendicularly on this same surface, let there be drawn a plane through the line CM and through KCH, making in the spheroid the semi-ellipse QM_q_, which will be given, since the angle MCL is given of value 6 degrees 40 minutes.

And it is certain, according to what has been explained above, Article 27, that a plane which would touch the spheroid at the point M, where I suppose the straight line CM to meet the surface, would be parallel to the plane QG_q_. If then through the point K one now draws KS parallel to G_g_, which will be parallel also to QX, the tangent to the Ellipse QG_q_ at Q; and if one conceives a plane pa.s.sing through KS and touching the spheroid, the point of contact will necessarily be in the Ellipse QM_q_, because this plane through KS, as well as the plane which touches the spheroid at the point M, are parallel to QX, the tangent of the spheroid: for this consequence will be demonstrated at the end of this Treatise. Let this point of contact be at I, then making KC, QC, DC proportionals, draw DI parallel to CM; also join CI.

I say that CI will be the required refraction of the ray RC. This will be manifest if, in considering CO, which is perpendicular to the ray RC, as a portion of the wave of light, we can demonstrate that the continuation of its piece C will be found in the crystal at I, when O has arrived at K.

38. Now as in the Chapter on Reflexion, in demonstrating that the incident and reflected rays are always in the same plane perpendicular to the reflecting surface, we considered the breadth of the wave of light, so, similarly, we must here consider the breadth of the wave CO in the diameter G_g_. Taking then the breadth C_c_ on the side toward the angle E, let the parallelogram CO_oc_ be taken as a portion of a wave, and let us complete the parallelograms CK_kc_, CI_ic_, Kl_ik_, OK_ko_. In the time then that the line O_o_ arrives at the surface of the crystal at K_k_, all the points of the wave CO_oc_ will have arrived at the rectangle K_c_ along lines parallel to OK; and from the points of their incidences there will originate, beyond that, in the crystal partial hemi-spheroids, similar to the hemi-spheroid QM_q_, and similarly disposed. These hemi-spheroids will necessarily all touch the plane of the parallelogram KI_ik_ at the same instant that O_o_ has reached K_k_. Which is easy to comprehend, since, of these hemi-spheroids, all those which have their centres along the line CK, touch this plane in the line KI (for this is to be shown in the same way as we have demonstrated the refraction of the oblique ray in the princ.i.p.al section through EF) and all those which have their centres in the line C_c_ will touch the same plane KI in the line I_i_; all these being similar to the hemi-spheroid QM_q_. Since then the parallelogram K_i_ is that which touches all these spheroids, this same parallelogram will be precisely the continuation of the wave CO_oc_ in the crystal, when O_o_ has arrived at K_k_, because it forms the termination of the movement and because of the quant.i.ty of movement which occurs more there than anywhere else: and thus it appears that the piece C of the wave CO_oc_ has its continuation at I; that is to say, that the ray RC is refracted as CI.

From this it is to be noted that the proportion of the refraction for this section of the crystal is that of the line N to the semi-diameter CQ; by which one will easily find the refractions of all incident rays, in the same way as we have shown previously for the case of the section through FE; and the demonstration will be the same. But it appears that the said proportion of the refraction is less here than in the section through FEB; for it was there the same as the ratio of N to CG, that is to say, as 156,962 to 98,779, very nearly as 8 to 5; and here it is the ratio of N to CQ the major semi-diameter of the spheroid, that is to say, as 156,962 to 105,032, very nearly as 3 to 2, but just a little less. Which still agrees perfectly with what one finds by observation.

39. For the rest, this diversity of proportion of refraction produces a very singular effect in this Crystal; which is that when it is placed upon a sheet of paper on which there are letters or anything else marked, if one views it from above with the two eyes situated in the plane of the section through EF, one sees the letters raised up by this irregular refraction more than when one puts one's eyes in the plane of section through AH: and the difference of these elevations appears by comparison with the other ordinary refraction of the crystal, the proportion of which is as 5 to 3, and which always raises the letters equally, and higher than the irregular refraction does.

For one sees the letters and the paper on which they are written, as on two different stages at the same time; and in the first position of the eyes, namely, when they are in the plane through AH these two stages are four times more distant from one another than when the eyes are in the plane through EF.

We will show that this effect follows from the refractions; and it will enable us at the same time to ascertain the apparent place of a point of an object placed immediately under the crystal, according to the different situation of the eyes.

40. Let us see first by how much the irregular refraction of the plane through AH ought to lift the bottom of the crystal. Let the plane of this figure represent separately the section through Q_q_ and CL, in which section there is also the ray RC, and let the semi-elliptic plane through Q_q_ and CM be inclined to the former, as previously, by an angle of 6 degrees 40 minutes; and in this plane CI is then the refraction of the ray RC.

[Ill.u.s.tration]

If now one considers the point I as at the bottom of the crystal, and that it is viewed by the rays ICR, _Icr_, refracted equally at the points C_c_, which should be equally distant from D, and that these rays meet the two eyes at R_r_; it is certain that the point I will appear raised to S where the straight lines RC, _rc_, meet; which point S is in DP, perpendicular to Q_q_. And if upon DP there is drawn the perpendicular IP, which will lie at the bottom of the crystal, the length SP will be the apparent elevation of the point I above the bottom.

Let there be described on Q_q_ a semicircle cutting the ray CR at B, from which BV is drawn perpendicular to Q_q_; and let the proportion of the refraction for this section be, as before, that of the line N to the semi-diameter CQ.

Then as N is to CQ so is VC to CD, as appears by the method of finding the refraction which we have shown above, Article 31; but as VC is to CD, so is VB to DS. Then as N is to CQ, so is VB to DS. Let ML be perpendicular to CL. And because I suppose the eyes R_r_ to be distant about a foot or so from the crystal, and consequently the angle RS_r_ very small, VB may be considered as equal to the semi-diameter CQ, and DP as equal to CL; then as N is to CQ so is CQ to DS. But N is valued at 156,962 parts, of which CM contains 100,000 and CQ 105,032. Then DS will have 70,283. But CL is 99,324, being the sine of the complement of the angle MCL which is 6 degrees 40 minutes; CM being supposed as radius. Then DP, considered as equal to CL, will be to DS as 99,324 to 70,283. And so the elevation of the point I by the refraction of this section is known.

[Ill.u.s.tration]

41. Now let there be represented the other section through EF in the figure before the preceding one; and let CM_g_ be the semi-ellipse, considered in Articles 27 and 28, which is made by cutting a spheroidal wave having centre C. Let the point I, taken in this ellipse, be imagined again at the bottom of the Crystal; and let it be viewed by the refracted rays ICR, I_cr_, which go to the two eyes; CR and _cr_ being equally inclined to the surface of the crystal G_g_.

This being so, if one draws ID parallel to CM, which I suppose to be the refraction of the perpendicular ray incident at the point C, the distances DC, D_c_, will be equal, as is easy to see by that which has been demonstrated in Article 28. Now it is certain that the point I should appear at S where the straight lines RC, _rc_, meet when prolonged; and that this point will fall in the line DP perpendicular to G_g_. If one draws IP perpendicular to this DP, it will be the distance PS which will mark the apparent elevation of the point I. Let there be described on G_g_ a semicircle cutting CR at B, from which let BV be drawn perpendicular to G_g_; and let N to GC be the proportion of the refraction in this section, as in Article 28. Since then CI is the refraction of the radius BC, and DI is parallel to CM, VC must be to CD as N to GC, according to what has been demonstrated in Article 31. But as VC is to CD so is BV to DS. Let ML be drawn perpendicular to CL. And because I consider, again, the eyes to be distant above the crystal, BV is deemed equal to the semi-diameter CG; and hence DS will be a third proportional to the lines N and CG: also DP will be deemed equal to CL. Now CG consisting of 98,778 parts, of which CM contains 100,000, N is taken as 156,962. Then DS will be 62,163. But CL is also determined, and contains 99,324 parts, as has been said in Articles 34 and 40. Then the ratio of PD to DS will be as 99,324 to 62,163. And thus one knows the elevation of the point at the bottom I by the refraction of this section; and it appears that this elevation is greater than that by the refraction of the preceding section, since the ratio of PD to DS was there as 99,324 to 70,283.

[Ill.u.s.tration]

But by the regular refraction of the crystal, of which we have above said that the proportion is 5 to 3, the elevation of the point I, or P, from the bottom, will be 2/5 of the height DP; as appears by this figure, where the point P being viewed by the rays PCR, P_cr_, refracted equally at the surface C_c_, this point must needs appear to be at S, in the perpendicular PD where the lines RC, _rc_, meet when prolonged: and one knows that the line PC is to CS as 5 to 3, since they are to one another as the sine of the angle CSP or DSC is to the sine of the angle SPC. And because the ratio of PD to DS is deemed the same as that of PC to CS, the two eyes Rr being supposed very far above the crystal, the elevation PS will thus be 2/5 of PD.

[Ill.u.s.tration]

42. If one takes a straight line AB for the thickness of the crystal, its point B being at the bottom, and if one divides it at the points C, D, E, according to the proportions of the elevations found, making AE 3/5 of AB, AB to AC as 99,324 to 70,283, and AB to AD as 99,324 to 62,163, these points will divide AB as in this figure. And it will be found that this agrees perfectly with experiment; that is to say by placing the eyes above in the plane which cuts the crystal according to the shorter diameter of the rhombus, the regular refraction will lift up the letters to E; and one will see the bottom, and the letters over which it is placed, lifted up to D by the irregular refraction.

But by placing the eyes above in the plane which cuts the crystal according to the longer diameter of the rhombus, the regular refraction will lift the letters to E as before; but the irregular refraction will make them, at the same time, appear lifted up only to C; and in such a way that the interval CE will be quadruple the interval ED, which one previously saw.

43. I have only to make the remark here that in both the positions of the eyes the images caused by the irregular refraction do not appear directly below those which proceed from the regular refraction, but they are separated from them by being more distant from the equilateral solid angle of the Crystal. That follows, indeed, from all that has been hitherto demonstrated about the irregular refraction; and it is particularly shown by these last demonstrations, from which one sees that the point I appears by irregular refraction at S in the perpendicular line DP, in which line also the image of the point P ought to appear by regular refraction, but not the image of the point I, which will be almost directly above the same point, and higher than S.

But as to the apparent elevation of the point I in other positions of the eyes above the crystal, besides the two positions which we have just examined, the image of that point by the irregular refraction will always appear between the two heights of D and C, pa.s.sing from one to the other as one turns one's self around about the immovable crystal, while looking down from above. And all this is still found conformable to our hypothesis, as any one can a.s.sure himself after I shall have shown here the way of finding the irregular refractions which appear in all other sections of the crystal, besides the two which we have considered. Let us suppose one of the faces of the crystal, in which let there be the Ellipse HDE, the centre C of which is also the centre of the spheroid HME in which the light spreads, and of which the said Ellipse is the section. And let the incident ray be RC, the refraction of which it is required to find.

Let there be taken a plane pa.s.sing through the ray RC and which is perpendicular to the plane of the ellipse HDE, cutting it along the straight line BCK; and having in the same plane through RC made CO perpendicular to CR, let OK be adjusted across the angle OCK, so as to be perpendicular to OC and equal to the line N, which I suppose to measure the travel of the light in air during the time that it spreads in the crystal through the spheroid HDEM. Then in the plane of the Ellipse HDE let KT be drawn, through the point K, perpendicular to BCK. Now if one conceives a plane drawn through the straight line KT and touching the spheroid HME at I, the straight line CI will be the refraction of the ray RC, as is easy to deduce from that which has been demonstrated in Article 36.

Treatise on Light Part 4

You're reading novel Treatise on Light Part 4 online at LightNovelFree.com. You can use the follow function to bookmark your favorite novel ( Only for registered users ). If you find any errors ( broken links, can't load photos, etc.. ), Please let us know so we can fix it as soon as possible. And when you start a conversation or debate about a certain topic with other people, please do not offend them just because you don't like their opinions.


Treatise on Light Part 4 summary

You're reading Treatise on Light Part 4. This novel has been translated by Updating. Author: Christiaan Huygens already has 579 views.

It's great if you read and follow any novel on our website. We promise you that we'll bring you the latest, hottest novel everyday and FREE.

LightNovelFree.com is a most smartest website for reading novel online, it can automatic resize images to fit your pc screen, even on your mobile. Experience now by using your smartphone and access to LightNovelFree.com